Question
For a particle in simple harmonic motion, show that $v_{\max }=$ $(\pi / 2) v_{\mathrm{av}_{2}}$ where $v_{\mathrm{arg}}$ is the average speed during one cycle of the motion.
Step 1
In this case, the displacement is the distance the particle moves in one half period of the motion, from $x = -a$ to $x = a$. Therefore, the displacement is $2a$. Show more…
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Maximum speed of a particle in simple harmonic motion is vmax. Then average speed of a particle in SHM is equal to:
An object moves with simple harmonic motion of period $T$ and amplitude $A .$ During one complete cycle, for what length of time is the speed of the object greater than $v_{max } / 2 ?$
An object moves with simple harmonic motion of period $T$ and amplitude $A .$ During one complete cycle, for what length of time is the speed of the object greater than $v_{\max } / 2 ?$
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