A body is performing S.H.M. Then its: - Vidyadip

MCQ

A body is performing $\text{S.H.M.}$ Then its:

A
Average total energy per cycle is equal to its maximum kinetic energy.
B
Average kinetic energy per cycle is equal to half of its maximum kinetic energy.
C
Root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity.
✓
All of the above

✓

Answer

Correct option: D.

All of the above

In case of $\text{S.H.M,}$ average total energy per cycle
$=$ Maximum kinetic energy $(K_0)$
$=$ Maximum potential energy $(U_0)$
Average $KE$ per cycle $=\frac{0+\text{K}_0}{2}=\frac{\text{K}_0}{2}$
Let us write the equation for the $\text{SHM} \text{x}=\text{a}\sin\omega\text{t}.$
Clearly, it is a periodic motion as it involves sine function.
Let us find velocity of the particle, $\text{v}=\frac{\text{dx}}{\text{dt}}=\frac{\text{d}}{\text{dt}}(\text{a}\sin\omega\text{t})=\text{a}\omega\cos\omega\text{t}$
Mean velocity over a complete cycle,
$\text{v}_\text{mean}=\frac{\int_{0}^{2\pi}\omega\text{a}\cos\theta\text{d}\theta}{2\pi}=\frac{\omega\text{a}[\sin\theta]^2_0}{2\pi}=0$
So, $\text{v}_\text{mean}\neq\frac{2}{\pi}\text{v}_\text{max}$
Root mean square speed,
$\text{v}_\text{ms}=\sqrt{\frac{\text{v}^2_\text{min}+\text{v}^2_\text{max}}{2}}=\sqrt{\frac{0+\text{v}^2-\text{max}}{2}}$
$\text{v}_\text{ms}=\frac{1}{\sqrt{2}}\text{v}_\text{max}$

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