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modulus
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Homework Statement
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?
Homework 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)
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 [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??