# Homework Help: Answer to an SHM Problem Does Not Fit Into Original Situation

1. Dec 11, 2012

### modulus

1. The problem statement, all variables and given/known data
A particle performs SHM of amplitude A along a straight line. When it is at a distance (√3/2)A from the mean position, its kinetic energy gets increased by an amount (0.5)mω2A2 due to an impulsive force. Then its new amplitude becomes?

2. Relevant equations
The basic idea is that the total energy (KE + PE) of a system in SHM remains constant.

Also of use:
v = Aω*cos(ωt)
vmax = Aω, in an SHM of angular frequency ω and amplitude A.
Kinetic Energy = 1/2(mv2)

3. The attempt at a solution
So, once the system gains the given amount of extra energy, it's total energy becomes mω2A'2.
Which means the amplitude A' will be equal to (√2)A (that is assuming that the value of ω does not change...which makes complete sense).

But, I run into a little problem when I try to use this result in the original situation:
I figured out that the speed of the particle at the position given (before the impulse) would be Aω/2. After the addition of kinetic energy given in the problem, the speed will become $\frac{(√5)Aω}{2}$. Comparing this with the first equation I've listed, that implies that A' = (√5)A (because ω doesn't change).

....What went wrong??

2. Dec 11, 2012

### Staff: Mentor

Prior to the impulse, the physical position of the mass corresponded to some angle ωt. That is,

$\frac{\sqrt{3}}{2}A = A sin(ω t)$

and angle θ = ωt turned out to be 60 degrees.

After the impulse adds energy to the system, the amplitude increases and this particular physical location no longer corresponds to an angle of 60 degrees in the cycle, so you can't assume that cos(ωt) will still be 1/2.

Edit: In order to maintain timeline continuity, the "new" equation for the position of the object would have to incorporate a phase angle so that the 'before' and 'after' equations meet seamlessly at the impulse instant. Suppose the impulse occurs at time tx, then

$x(t) = Asin(ω t) ~~~~~~~~~~~~~~0 \le t < t_x$
$x(t) = A'sin(ω t + \phi)~~~~~~~~~~t \ge t_x$

Last edited: Dec 11, 2012
3. Dec 11, 2012

### Staff: Mentor

The kinetic energy at that point is 1/8 m ω^2 A^2. You increase it by 1/2 m ω^2A^2, so you increase its velocity by a factor of sqrt(5).

The position where the particle gets its kick is $\frac{\sqrt{3}{2}}$ of the initial amplitude, this is $\sqrt\frac{3}{8}}$ of the new amplitude (using A'=√2*A). There, the particle has 5/8 of its total energy as kinetic energy and its velocity is the value you calculated.

Short version:
(√3/2)A != (√3/2)A'

4. Dec 11, 2012

### modulus

OK, so this is where I was going wrong:

A'ω cos(new phase angle) = ((√5)/2)Aω

And the "new phase angle" will adjust itself so that this relation is true for A' = A√2.
In fact, the "new phase angle" will be such that:

A' sin(new phase angle) = ((√3)/2)A

Oh...clarity!
Thanks, guys. :)