First slide

Simple harmonic motion

Question

At what displacement is the KE of a particle performing SHM of amplitude 10 cm equal to three times it’s PE?

Moderate

A

10 cm
B

5 cm
C

15 cm
D

20 cm

Solution

$we know KE= \frac{1}{2} {mw}^{2} (A^{2} {-x}^{2}), here m= mass of oscillator,w=angular velocity,A=amplitude,x=displacement P E = \frac{1}{2} {mw}^{2} (x^{2}) g i v e n K E = 3 P E, S u b s t i t u t e t h e a b o v e e q u a t i o n s \frac{1}{2} {mw}^{2} (A^{2} {-x}^{2}) = \frac{3}{2} {mw}^{2} (x^{2}) \Rightarrow A^{2} {=4x}^{2} \Rightarrow x=A/2 G i v e n A = 10 c m ∴ x = \frac{10}{2} = 5cm$

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