(adsbygoogle = window.adsbygoogle || []).push({}); 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)mω^{2}A^{2}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)

v_{max}= Aω, in an SHM of angular frequency ω and amplitude A.

Kinetic Energy = 1/2(mv^{2})

3. The attempt at a solution

So, once the system gains the given amount of extra energy, it's total energy becomes mω^{2}A'^{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 [itex]\frac{(√5)Aω}{2}[/itex]. Comparing this with the first equation I've listed, that implies that A' = (√5)A (because ω doesn't change).

....What went wrong??

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

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