First slide

Simple harmonic motion

Question

In a simple harmonic oscillator, at the mean position is

Moderate

A

KE maximum, PE maximum
B

KE is minimum, PE is maximum
C

KE is maximum, PE is minimum
D

both KE and PE are minimum

Solution

We know $KE= \frac{1}{2} {mω}^{2} (A^{2} {-x}^{2})$
$PE= \frac{1}{2} {mω}^{2} x^{2}, here m=mass of oscillator, ω=angular frequency,A=amplitude,x=displacement$
At mean position x=0
$KE= \frac{1}{2} {mω}^{2} A^{2}$ (maximum)
$PE=0$ (minimum).
**** For a simple harmonic oscillator at mean position potential energy is minimum but not equal to zero.
Its total energy =min P.E + max K.E=max P.E

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