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giveortake

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- Homework Statement
- A Spring is attached to a spring is released from rest 1 m below the equilibrium position for the mass–spring system and begins to vibrate. After π/2 sec, the mass is struck by a hammer exerting an impulse on the mass. The system is governed by the symbolic initial value problem

y'' + 9y =−3δ(t−π/2) where y(0)=1 and y'(0)=0. What does the mass do after the hammer strike?

- Relevant Equations
- Laplace Transforms:

L(y'') = s[SUP]2[/SUP]L - sy(0) - y'(0)

L(y) = L

1.) Laplace transform of differential equation, where L is the Laplace transform of y:

s

= s

2.) Solve for L

L = (-3e

3.) Solve for y by performing the inverse Laplace on L

Decompose L into 2 parts:

L = -3e

The inverse Laplace of -3e

-3⋅H(t-π/2)cos(3(t+π/2))

Where H is the Heaviside function.

The inverse Laplace of s/(s

cos(3t)

Thus y = -3⋅H(t-π/2)cos(3(t+π/2)) + cos(3t)

The spring-mass system follows this equation for simple harmonic motion after the hammer strike.

NOTE: The answer according to the back of the textbook is:

y = 0 for t > π/2

s

^{2}L - sy(0) - y'(0) + 9L = -3e^{-πs/2}= s

^{2}L - s+ 9L = -3e^{-πs/2}2.) Solve for L

L = (-3e

^{-πs/2}+ s) / (s^{2}+ 9)3.) Solve for y by performing the inverse Laplace on L

Decompose L into 2 parts:

L = -3e

^{-πs/2}/(s^{2}+ 9) + s/(s^{2}+ 9)The inverse Laplace of -3e

^{-πs/2}/(s^{2}+ 9) is:-3⋅H(t-π/2)cos(3(t+π/2))

Where H is the Heaviside function.

The inverse Laplace of s/(s

^{2}+ 9) is:cos(3t)

Thus y = -3⋅H(t-π/2)cos(3(t+π/2)) + cos(3t)

The spring-mass system follows this equation for simple harmonic motion after the hammer strike.

NOTE: The answer according to the back of the textbook is:

y = 0 for t > π/2