Dirac Delta Function in an Ordinary Differential Equation

In summary: 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/2I 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=
  • #1
giveortake
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0
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
 
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  • #2
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.
 
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  • #3
giveortake said:
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}##.
 
  • #4
giveortake said:
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.
 

Related to Dirac Delta Function in an Ordinary Differential Equation

1. What is the Dirac Delta Function?

The Dirac Delta Function is a mathematical function that is often used in physics and engineering to model impulsive forces or point sources. It is a generalized function that is defined as zero for all values except for one point, where it is infinite. It is represented by the symbol δ and is also known as the unit impulse function.

2. How is the Dirac Delta Function used in ordinary differential equations?

The Dirac Delta Function is often used in ordinary differential equations to represent a point source or impulse force. It is commonly used in equations that involve discontinuous or impulsive forces, such as in mechanics or circuit analysis problems. The Dirac Delta Function allows for the simplification of these equations and makes them easier to solve.

3. What is the relationship between the Dirac Delta Function and the Kronecker Delta?

The Dirac Delta Function and the Kronecker Delta are both mathematical functions that are used to represent a point or impulse. However, they are defined differently and have different properties. The Dirac Delta Function is a continuous function while the Kronecker Delta is a discrete function. The Dirac Delta Function is also often used in calculus and integration, while the Kronecker Delta is commonly used in discrete mathematics and linear algebra.

4. Can the Dirac Delta Function be integrated?

Yes, the Dirac Delta Function can be integrated. However, its integral is not a regular function and it is often defined using the concept of distributions in mathematics. The integral of the Dirac Delta Function is equal to 1.

5. Are there any real-life applications of the Dirac Delta Function?

Yes, the Dirac Delta Function has many real-life applications in physics and engineering. It is commonly used in solving problems related to mechanics, electricity and magnetism, and signal processing. It is also used in probability and statistics to model random events and in image processing to sharpen images and filter out noise.

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