Dirac Delta Function in an Ordinary Differential Equation

[giveortake]

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:

s²L - sy(0) - y'(0) + 9L = -3e^-πs/2

= s²L - s+ 9L = -3e^-πs/2

2.) Solve for L

L = (-3e^-πs/2 + s) / (s² + 9)

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

Decompose L into 2 parts:
L = -3e^-πs/2/(s² + 9) + s/(s² + 9)

The inverse Laplace of -3e^-πs/2/(s² + 9) is:

-3⋅H(t-π/2)cos(3(t+π/2))
Where H is the Heaviside function.

The inverse Laplace of s/(s² + 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

 

[DEvens]

I hope you made some typos in the statement of the problem. Maybe you could fix that?

Anyway, the claim that y=0 for t>pi/2 would indicate that the mass sits there at y=0. That is, it seems like the answer in the back of the book is saying the mass comes to rest. You could check that by checking some things.
- Is the mass at y=0 when the hammer hits it? If not, then the y=0 solution is wrong.
- If it is at y=0, then what is the velocity of the mass before it gets hit?
- Is the impulse the correct amount to bring the mass to rest? If not, then the y=0 solution is wrong.

If those things come out correctly to have the mass come to rest, then you don't really need to solve the full differential equation, only the before-getting-hit part.

 

[tnich]

L = -3e^-πs/2/(s² + 9) + s/(s² + 9)

There are several substitutions you need to make to find the inverse LaPlace transform of this, and I think you have applied one of them incorrectly: $sin(\omega t) \rightarrow \frac {\omega}{s^2+\omega^2}$.

 

[rude man]

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?
Homework Equations: Laplace Transforms:

L(y'') = s²L - sy(0) - y'(0)
L(y) = L

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

s²L - sy(0) - y'(0) + 9L = -3e^-πs/2

= s²L - s+ 9L = -3e^-πs/2

2.) Solve for L

L = (-3e^-πs/2 + s) / (s² + 9)

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

Decompose L into 2 parts:
L = -3e^-πs/2/(s² + 9) + s/(s² + 9)

OK so far

The inverse Laplace of -3e^-πs/2/(s² + 9) is:

-3⋅H(t-π/2)cos(3(t+π/2))

2 mistakes here:
Whence the "3" coefficient? (already pointed out in a previous reply)
And look again at the cosine argument.
If you fix this term you will get zero for $t \geq \pi/2$.
The key operation is $e^{-bs} F(s) \leftrightarrow f(t-b)H(t-b)$ where $f(t) \leftrightarrow F(s).$
I think you used $e^{-bs} F(s+b) \leftrightarrow f(t)H(t-b)$ which is not the same thing.