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

MCQ

A body is performing 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
Mean velocity over a complete cycle is equal to $\frac{2}{\pi}$ times of its maximum velocity.
D
Root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity.

✓

Answer

Average total energy per cycle is equal to its maximum kinetic energy.

Average kinetic energy per cycle is equal to half of its maximum kinetic energy.

Root mean square velocity is $\frac{1}{\sqrt{2}}$ times of its maximum velocity.

Explanation:

In case of S.H.M, average total energy per cycle

= Maximum kinetic energy (K₀)

= Maximum potential energy (U₀)

Average KE per cycle $=\frac{0+\text{K}_0}{2}=\frac{\text{K}_0}{2}$

Let us write the equation for the 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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