## S2m: Coupled mass-spring systems (ver. 1)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$15$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$5$$ meters inwards from its natural position, while the outer mass is moved $$\frac{29}{4}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 19 x_1+ 4 x_2\hspace{2em}x_1(0)= -5 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -\frac{29}{4} ,x_2'(0)=0$

This system solves to:

$x_1= -3 \, \cos\left(2 \, \sqrt{5} t\right) - 2 \, \cos\left(\sqrt{3} t\right)$

$x_2= \frac{3}{4} \, \cos\left(2 \, \sqrt{5} t\right) - 8 \, \cos\left(\sqrt{3} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-2.91$$ meters from its natural position, and the outer mass is located approximately $$-3.23$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 2)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$4$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$6$$ meters outwards from its natural position, while the outer mass is moved $$3$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$2 x_1''=- 6 x_1+ 2 x_2\hspace{2em}x_1(0)= 6 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 3 ,x_2'(0)=0$

This system solves to:

$x_1= 3 \, \cos\left(2 \, t\right) + 3 \, \cos\left(t\right)$

$x_2= -3 \, \cos\left(2 \, t\right) + 6 \, \cos\left(t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$-1.67$$ meters from its natural position, and the outer mass is located approximately $$4.22$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 3)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$15$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$10$$ meters outwards from its natural position, while the outer mass is moved $$\frac{75}{4}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 19 x_1+ 4 x_2\hspace{2em}x_1(0)= 10 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= \frac{75}{4} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(2 \, \sqrt{5} t\right) + 5 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{5}{4} \, \cos\left(2 \, \sqrt{5} t\right) + 20 \, \cos\left(\sqrt{3} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$5.63$$ meters from its natural position, and the outer mass is located approximately $$8.48$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 4)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$18$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$2$$ meters outwards from its natural position, while the outer mass is moved $$10$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 4 x_2\hspace{2em}x_1(0)= 2 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -10 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(2 \, \sqrt{3} t\right) - 2 \, \cos\left(\sqrt{3} t\right)$

$x_2= -2 \, \cos\left(2 \, \sqrt{3} t\right) - 8 \, \cos\left(\sqrt{3} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-0.490$$ meters from its natural position, and the outer mass is located approximately $$-6.95$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 5)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$18$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$1$$ meters inwards from its natural position, while the outer mass is moved $$14$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 4 x_2\hspace{2em}x_1(0)= -1 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= 14 ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(2 \, \sqrt{3} t\right) + 3 \, \cos\left(\sqrt{3} t\right)$

$x_2= 2 \, \cos\left(2 \, \sqrt{3} t\right) + 12 \, \cos\left(\sqrt{3} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$-6.04$$ meters from its natural position, and the outer mass is located approximately $$-9.78$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 6)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$0$$ meters outwards from its natural position, while the outer mass is moved $$5$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= 0 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 5 ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(\sqrt{6} t\right) + 2 \, \cos\left(t\right)$

$x_2= \cos\left(\sqrt{6} t\right) + 4 \, \cos\left(t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$0.555$$ meters from its natural position, and the outer mass is located approximately $$-3.55$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 7)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$8$$ meters inwards from its natural position, while the outer mass is moved $$\frac{17}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= -8 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -\frac{17}{2} ,x_2'(0)=0$

This system solves to:

$x_1= -3 \, \cos\left(\sqrt{6} t\right) - 5 \, \cos\left(t\right)$

$x_2= \frac{3}{2} \, \cos\left(\sqrt{6} t\right) - 10 \, \cos\left(t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$-4.27$$ meters from its natural position, and the outer mass is located approximately $$-1.41$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 8)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$\frac{20}{3}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{20}{3} ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(\sqrt{21} t\right) + 2 \, \cos\left(\sqrt{2} t\right)$

$x_2= \frac{2}{3} \, \cos\left(\sqrt{21} t\right) + 6 \, \cos\left(\sqrt{2} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$1.96$$ meters from its natural position, and the outer mass is located approximately $$-6.35$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 9)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$9$$ meters outwards from its natural position, while the outer mass is moved $$8$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= 9 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 8 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(\sqrt{6} t\right) + 5 \, \cos\left(t\right)$

$x_2= -2 \, \cos\left(\sqrt{6} t\right) + 10 \, \cos\left(t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-6.99$$ meters from its natural position, and the outer mass is located approximately $$-4.67$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 10)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$15$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$\frac{33}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 19 x_1+ 4 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -\frac{33}{2} ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(2 \, \sqrt{5} t\right) - 4 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{1}{2} \, \cos\left(2 \, \sqrt{5} t\right) - 16 \, \cos\left(\sqrt{3} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-0.540$$ meters from its natural position, and the outer mass is located approximately $$-7.77$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 11)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$7$$ meters inwards from its natural position, while the outer mass is moved $$2$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= -7 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= 2 ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(2 \, \sqrt{5} t\right) - 3 \, \cos\left(\sqrt{3} t\right)$

$x_2= 6 \, \cos\left(2 \, \sqrt{5} t\right) - 4 \, \cos\left(\sqrt{3} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$6.39$$ meters from its natural position, and the outer mass is located approximately $$-1.53$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 12)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$8$$ meters outwards from its natural position, while the outer mass is moved $$\frac{34}{3}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= 8 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{34}{3} ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(\sqrt{21} t\right) + 4 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{2}{3} \, \cos\left(\sqrt{21} t\right) + 12 \, \cos\left(\sqrt{2} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$-7.67$$ meters from its natural position, and the outer mass is located approximately $$-10.8$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 13)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$6$$ meters outwards from its natural position, while the outer mass is moved $$\frac{15}{2}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 9 x_1+ 4 x_2\hspace{2em}x_1(0)= 6 ,x_1'(0)=0$

$2 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= \frac{15}{2} ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(\sqrt{10} t\right) + 4 \, \cos\left(t\right)$

$x_2= -\frac{1}{2} \, \cos\left(\sqrt{10} t\right) + 8 \, \cos\left(t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$0.334$$ meters from its natural position, and the outer mass is located approximately $$-3.83$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 14)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$0$$ meters outwards from its natural position, while the outer mass is moved $$9$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 9 x_1+ 4 x_2\hspace{2em}x_1(0)= 0 ,x_1'(0)=0$

$2 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -9 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(\sqrt{10} t\right) - 4 \, \cos\left(t\right)$

$x_2= -\cos\left(\sqrt{10} t\right) - 8 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-0.0322$$ meters from its natural position, and the outer mass is located approximately $$8.92$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 15)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$\frac{25}{3}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= -\frac{25}{3} ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(2 \, \sqrt{5} t\right) - 4 \, \cos\left(\sqrt{3} t\right)$

$x_2= -3 \, \cos\left(2 \, \sqrt{5} t\right) - \frac{16}{3} \, \cos\left(\sqrt{3} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-0.540$$ meters from its natural position, and the outer mass is located approximately $$-4.46$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 16)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$15$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$\frac{33}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 19 x_1+ 4 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -\frac{33}{2} ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(2 \, \sqrt{5} t\right) - 4 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{1}{2} \, \cos\left(2 \, \sqrt{5} t\right) - 16 \, \cos\left(\sqrt{3} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-2.05$$ meters from its natural position, and the outer mass is located approximately $$-13.1$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 17)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$4$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$3$$ meters outwards from its natural position, while the outer mass is moved $$9$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$2 x_1''=- 6 x_1+ 2 x_2\hspace{2em}x_1(0)= 3 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -9 ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(2 \, t\right) - 2 \, \cos\left(t\right)$

$x_2= -5 \, \cos\left(2 \, t\right) - 4 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$6.78$$ meters from its natural position, and the outer mass is located approximately $$-0.841$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 18)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$5$$ meters outwards from its natural position, while the outer mass is moved $$\frac{13}{4}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 9 x_1+ 4 x_2\hspace{2em}x_1(0)= 5 ,x_1'(0)=0$

$2 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= \frac{13}{4} ,x_2'(0)=0$

This system solves to:

$x_1= 3 \, \cos\left(\sqrt{10} t\right) + 2 \, \cos\left(t\right)$

$x_2= -\frac{3}{4} \, \cos\left(\sqrt{10} t\right) + 4 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-4.97$$ meters from its natural position, and the outer mass is located approximately $$-3.21$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 19)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$2$$ meters outwards from its natural position, while the outer mass is moved $$\frac{47}{6}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 11 x_1+ 6 x_2\hspace{2em}x_1(0)= 2 ,x_1'(0)=0$

$1 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= -\frac{47}{6} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(\sqrt{15} t\right) - 3 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{10}{3} \, \cos\left(\sqrt{15} t\right) - \frac{9}{2} \, \cos\left(\sqrt{2} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-7.31$$ meters from its natural position, and the outer mass is located approximately $$-0.390$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 20)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$8$$ meters outwards from its natural position, while the outer mass is moved $$\frac{10}{3}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 11 x_1+ 6 x_2\hspace{2em}x_1(0)= 8 ,x_1'(0)=0$

$1 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{10}{3} ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(\sqrt{15} t\right) + 4 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{8}{3} \, \cos\left(\sqrt{15} t\right) + 6 \, \cos\left(\sqrt{2} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-0.667$$ meters from its natural position, and the outer mass is located approximately $$7.47$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 21)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$1$$ meters outwards from its natural position, while the outer mass is moved $$\frac{43}{6}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= 1 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= -\frac{43}{6} ,x_2'(0)=0$

This system solves to:

$x_1= 3 \, \cos\left(2 \, \sqrt{5} t\right) - 2 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{9}{2} \, \cos\left(2 \, \sqrt{5} t\right) - \frac{8}{3} \, \cos\left(\sqrt{3} t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$-1.35$$ meters from its natural position, and the outer mass is located approximately $$6.12$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 22)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$1$$ meters outwards from its natural position, while the outer mass is moved $$\frac{25}{2}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= 1 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{25}{2} ,x_2'(0)=0$

This system solves to:

$x_1= -3 \, \cos\left(\sqrt{21} t\right) + 4 \, \cos\left(\sqrt{2} t\right)$

$x_2= \frac{1}{2} \, \cos\left(\sqrt{21} t\right) + 12 \, \cos\left(\sqrt{2} t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$4.64$$ meters from its natural position, and the outer mass is located approximately $$8.16$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 23)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$1$$ meters outwards from its natural position, while the outer mass is moved $$\frac{25}{2}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= 1 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{25}{2} ,x_2'(0)=0$

This system solves to:

$x_1= -3 \, \cos\left(\sqrt{21} t\right) + 4 \, \cos\left(\sqrt{2} t\right)$

$x_2= \frac{1}{2} \, \cos\left(\sqrt{21} t\right) + 12 \, \cos\left(\sqrt{2} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-2.95$$ meters from its natural position, and the outer mass is located approximately $$-5.24$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 24)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$7$$ meters outwards from its natural position, while the outer mass is moved $$\frac{31}{6}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= 7 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{31}{6} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(\sqrt{21} t\right) + 2 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{5}{6} \, \cos\left(\sqrt{21} t\right) + 6 \, \cos\left(\sqrt{2} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$5.96$$ meters from its natural position, and the outer mass is located approximately $$4.14$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 25)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$4$$ meters inwards from its natural position, while the outer mass is moved $$\frac{17}{3}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= -4 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= -\frac{17}{3} ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(\sqrt{21} t\right) - 2 \, \cos\left(\sqrt{2} t\right)$

$x_2= \frac{1}{3} \, \cos\left(\sqrt{21} t\right) - 6 \, \cos\left(\sqrt{2} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-3.36$$ meters from its natural position, and the outer mass is located approximately $$-4.57$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 26)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$4$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$5$$ meters inwards from its natural position, while the outer mass is moved $$4$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$2 x_1''=- 6 x_1+ 2 x_2\hspace{2em}x_1(0)= -5 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -4 ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(2 \, t\right) - 3 \, \cos\left(t\right)$

$x_2= 2 \, \cos\left(2 \, t\right) - 6 \, \cos\left(t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$0.827$$ meters from its natural position, and the outer mass is located approximately $$-3.38$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 27)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$6$$ meters inwards from its natural position, while the outer mass is moved $$\frac{35}{3}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= -6 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= -\frac{35}{3} ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(\sqrt{21} t\right) - 4 \, \cos\left(\sqrt{2} t\right)$

$x_2= \frac{1}{3} \, \cos\left(\sqrt{21} t\right) - 12 \, \cos\left(\sqrt{2} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-4.98$$ meters from its natural position, and the outer mass is located approximately $$-9.43$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 28)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$8$$ meters inwards from its natural position, while the outer mass is moved $$\frac{7}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= -8 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -\frac{7}{2} ,x_2'(0)=0$

This system solves to:

$x_1= -5 \, \cos\left(\sqrt{6} t\right) - 3 \, \cos\left(t\right)$

$x_2= \frac{5}{2} \, \cos\left(\sqrt{6} t\right) - 6 \, \cos\left(t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$0.321$$ meters from its natural position, and the outer mass is located approximately $$2.96$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 29)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$6$$ meters outwards from its natural position, while the outer mass is moved $$3$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 9 x_1+ 4 x_2\hspace{2em}x_1(0)= 6 ,x_1'(0)=0$

$2 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= 3 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(\sqrt{10} t\right) + 2 \, \cos\left(t\right)$

$x_2= -\cos\left(\sqrt{10} t\right) + 4 \, \cos\left(t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$3.16$$ meters from its natural position, and the outer mass is located approximately $$-2.66$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 30)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$15$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$6$$ meters inwards from its natural position, while the outer mass is moved $$\frac{31}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$1 x_1''=- 19 x_1+ 4 x_2\hspace{2em}x_1(0)= -6 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -\frac{31}{2} ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(2 \, \sqrt{5} t\right) - 4 \, \cos\left(\sqrt{3} t\right)$

$x_2= \frac{1}{2} \, \cos\left(2 \, \sqrt{5} t\right) - 16 \, \cos\left(\sqrt{3} t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$4.75$$ meters from its natural position, and the outer mass is located approximately $$11.1$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 31)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$2$$ meters outwards from its natural position, while the outer mass is moved $$\frac{47}{6}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 11 x_1+ 6 x_2\hspace{2em}x_1(0)= 2 ,x_1'(0)=0$

$1 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= -\frac{47}{6} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(\sqrt{15} t\right) - 3 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{10}{3} \, \cos\left(\sqrt{15} t\right) - \frac{9}{2} \, \cos\left(\sqrt{2} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-7.31$$ meters from its natural position, and the outer mass is located approximately $$-0.390$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 32)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$6$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 6 ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(\sqrt{6} t\right) + 2 \, \cos\left(t\right)$

$x_2= 2 \, \cos\left(\sqrt{6} t\right) + 4 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-3.92$$ meters from its natural position, and the outer mass is located approximately $$-2.99$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 33)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$4$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$8$$ meters inwards from its natural position, while the outer mass is moved $$7$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$2 x_1''=- 6 x_1+ 2 x_2\hspace{2em}x_1(0)= -8 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -7 ,x_2'(0)=0$

This system solves to:

$x_1= -3 \, \cos\left(2 \, t\right) - 5 \, \cos\left(t\right)$

$x_2= 3 \, \cos\left(2 \, t\right) - 10 \, \cos\left(t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$3.70$$ meters from its natural position, and the outer mass is located approximately $$6.10$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 34)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$6$$ meters inwards from its natural position, while the outer mass is moved $$7$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= -6 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -7 ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(\sqrt{6} t\right) - 4 \, \cos\left(t\right)$

$x_2= \cos\left(\sqrt{6} t\right) - 8 \, \cos\left(t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$1.29$$ meters from its natural position, and the outer mass is located approximately $$3.51$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 35)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$1$$ meters outwards from its natural position, while the outer mass is moved $$\frac{19}{2}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= 1 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= \frac{19}{2} ,x_2'(0)=0$

This system solves to:

$x_1= -3 \, \cos\left(\sqrt{6} t\right) + 4 \, \cos\left(t\right)$

$x_2= \frac{3}{2} \, \cos\left(\sqrt{6} t\right) + 8 \, \cos\left(t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$-1.71$$ meters from its natural position, and the outer mass is located approximately $$3.69$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 36)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$4$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$6$$ meters outwards from its natural position, while the outer mass is moved $$0$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$2 x_1''=- 6 x_1+ 2 x_2\hspace{2em}x_1(0)= 6 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 0 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(2 \, t\right) + 2 \, \cos\left(t\right)$

$x_2= -4 \, \cos\left(2 \, t\right) + 4 \, \cos\left(t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-1.89$$ meters from its natural position, and the outer mass is located approximately $$-2.03$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 37)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$1$$ meters inwards from its natural position, while the outer mass is moved $$7$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 9 x_1+ 4 x_2\hspace{2em}x_1(0)= -1 ,x_1'(0)=0$

$2 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= 7 ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(\sqrt{10} t\right) + 3 \, \cos\left(t\right)$

$x_2= \cos\left(\sqrt{10} t\right) + 6 \, \cos\left(t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-5.95$$ meters from its natural position, and the outer mass is located approximately $$-2.93$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 38)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$18$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$4$$ meters outwards from its natural position, while the outer mass is moved $$7$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 4 x_2\hspace{2em}x_1(0)= 4 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= 7 ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(2 \, \sqrt{3} t\right) + 2 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\cos\left(2 \, \sqrt{3} t\right) + 8 \, \cos\left(\sqrt{3} t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$-1.36$$ meters from its natural position, and the outer mass is located approximately $$-5.81$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 39)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$3$$ meters outwards from its natural position, while the outer mass is moved $$\frac{61}{6}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= 3 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= -\frac{61}{6} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(2 \, \sqrt{5} t\right) - 2 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{15}{2} \, \cos\left(2 \, \sqrt{5} t\right) - \frac{8}{3} \, \cos\left(\sqrt{3} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$-2.54$$ meters from its natural position, and the outer mass is located approximately $$9.18$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 40)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$15$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$8$$ meters outwards from its natural position, while the outer mass is moved $$15$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$1 x_1''=- 19 x_1+ 4 x_2\hspace{2em}x_1(0)= 8 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= 15 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(2 \, \sqrt{5} t\right) + 4 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\cos\left(2 \, \sqrt{5} t\right) + 16 \, \cos\left(\sqrt{3} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$-7.34$$ meters from its natural position, and the outer mass is located approximately $$-14.3$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 41)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$6$$ meters outwards from its natural position, while the outer mass is moved $$7$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= 6 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 7 ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(\sqrt{6} t\right) + 4 \, \cos\left(t\right)$

$x_2= -\cos\left(\sqrt{6} t\right) + 8 \, \cos\left(t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-4.48$$ meters from its natural position, and the outer mass is located approximately $$-4.30$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 42)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$1$$ meters inwards from its natural position, while the outer mass is moved $$10$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$5$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= -1 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= 10 ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(2 \, \sqrt{5} t\right) + 3 \, \cos\left(\sqrt{3} t\right)$

$x_2= 6 \, \cos\left(2 \, \sqrt{5} t\right) + 4 \, \cos\left(\sqrt{3} t\right)$

Thus after $$5$$ seconds, the inner mass is located approximately $$1.56$$ meters from its natural position, and the outer mass is located approximately $$-8.48$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 43)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$3$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$1$$ meters outwards from its natural position, while the outer mass is moved $$\frac{11}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 5 x_1+ 2 x_2\hspace{2em}x_1(0)= 1 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= -\frac{11}{2} ,x_2'(0)=0$

This system solves to:

$x_1= 3 \, \cos\left(\sqrt{6} t\right) - 2 \, \cos\left(t\right)$

$x_2= -\frac{3}{2} \, \cos\left(\sqrt{6} t\right) - 4 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$3.43$$ meters from its natural position, and the outer mass is located approximately $$3.23$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 44)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$5$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 9 x_1+ 4 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$2 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= 5 ,x_2'(0)=0$

This system solves to:

$x_1= -4 \, \cos\left(\sqrt{10} t\right) + 2 \, \cos\left(t\right)$

$x_2= \cos\left(\sqrt{10} t\right) + 4 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$2.01$$ meters from its natural position, and the outer mass is located approximately $$-4.96$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 45)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$5$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$2$$ meters inwards from its natural position, while the outer mass is moved $$\frac{22}{3}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 11 x_1+ 6 x_2\hspace{2em}x_1(0)= -2 ,x_1'(0)=0$

$1 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= -\frac{22}{3} ,x_2'(0)=0$

This system solves to:

$x_1= 2 \, \cos\left(\sqrt{15} t\right) - 4 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{4}{3} \, \cos\left(\sqrt{15} t\right) - 6 \, \cos\left(\sqrt{2} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$2.98$$ meters from its natural position, and the outer mass is located approximately $$1.94$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 46)

Consider a coupled mass-spring system where the inner mass is $$1$$ kg, the outer mass is $$2$$ kg, the inner spring has constant $$14$$ N/m, the outer spring has constant $$6$$ N/m. The inner mass is moved $$7$$ meters outwards from its natural position, while the outer mass is moved $$\frac{25}{3}$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$1 x_1''=- 20 x_1+ 6 x_2\hspace{2em}x_1(0)= 7 ,x_1'(0)=0$

$2 x_2''= 6 x_1- 6 x_2\hspace{2em}x_2(0)= \frac{25}{3} ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(\sqrt{21} t\right) + 3 \, \cos\left(\sqrt{2} t\right)$

$x_2= -\frac{2}{3} \, \cos\left(\sqrt{21} t\right) + 9 \, \cos\left(\sqrt{2} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$0.160$$ meters from its natural position, and the outer mass is located approximately $$-4.33$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 47)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$1$$ meters outwards from its natural position, while the outer mass is moved $$10$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$2$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= 1 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= -10 ,x_2'(0)=0$

This system solves to:

$x_1= 4 \, \cos\left(2 \, \sqrt{5} t\right) - 3 \, \cos\left(\sqrt{3} t\right)$

$x_2= -6 \, \cos\left(2 \, \sqrt{5} t\right) - 4 \, \cos\left(\sqrt{3} t\right)$

Thus after $$2$$ seconds, the inner mass is located approximately $$-0.702$$ meters from its natural position, and the outer mass is located approximately $$9.11$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 48)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$10$$ N/m, the outer spring has constant $$12$$ N/m. The inner mass is moved $$3$$ meters outwards from its natural position, while the outer mass is moved $$\frac{61}{6}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 12 x_2\hspace{2em}x_1(0)= 3 ,x_1'(0)=0$

$1 x_2''= 12 x_1- 12 x_2\hspace{2em}x_2(0)= -\frac{61}{6} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(2 \, \sqrt{5} t\right) - 2 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{15}{2} \, \cos\left(2 \, \sqrt{5} t\right) - \frac{8}{3} \, \cos\left(\sqrt{3} t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$2.37$$ meters from its natural position, and the outer mass is located approximately $$-6.19$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 49)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$4$$ N/m, the outer spring has constant $$2$$ N/m. The inner mass is moved $$3$$ meters outwards from its natural position, while the outer mass is moved $$12$$ meters outwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$3$$ seconds.

The IVP system is given by:

$2 x_1''=- 6 x_1+ 2 x_2\hspace{2em}x_1(0)= 3 ,x_1'(0)=0$

$1 x_2''= 2 x_1- 2 x_2\hspace{2em}x_2(0)= 12 ,x_2'(0)=0$

This system solves to:

$x_1= -2 \, \cos\left(2 \, t\right) + 5 \, \cos\left(t\right)$

$x_2= 2 \, \cos\left(2 \, t\right) + 10 \, \cos\left(t\right)$

Thus after $$3$$ seconds, the inner mass is located approximately $$-6.87$$ meters from its natural position, and the outer mass is located approximately $$-7.98$$ meters from its natural position.

## S2m: Coupled mass-spring systems (ver. 50)

Consider a coupled mass-spring system where the inner mass is $$2$$ kg, the outer mass is $$1$$ kg, the inner spring has constant $$18$$ N/m, the outer spring has constant $$4$$ N/m. The inner mass is moved $$0$$ meters outwards from its natural position, while the outer mass is moved $$\frac{45}{2}$$ meters inwards from its natural position. Both masses are then simultaneously released from rest.

Give a system of IVPs that models this scenario, then solve the system. Use your solution to find the position of both masses after $$4$$ seconds.

The IVP system is given by:

$2 x_1''=- 22 x_1+ 4 x_2\hspace{2em}x_1(0)= 0 ,x_1'(0)=0$

$1 x_2''= 4 x_1- 4 x_2\hspace{2em}x_2(0)= -\frac{45}{2} ,x_2'(0)=0$

This system solves to:

$x_1= 5 \, \cos\left(2 \, \sqrt{3} t\right) - 5 \, \cos\left(\sqrt{3} t\right)$

$x_2= -\frac{5}{2} \, \cos\left(2 \, \sqrt{3} t\right) - 20 \, \cos\left(\sqrt{3} t\right)$

Thus after $$4$$ seconds, the inner mass is located approximately $$-2.61$$ meters from its natural position, and the outer mass is located approximately $$-16.7$$ meters from its natural position.