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A body is performing SHM, 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

Answers (1)

The answer is the option (a) Average total energy per cycle is equal to its maximum kinetic energy, (b) Average kinetic energy per cycle is equal to half its maximum kinetic energy, and (d) Root square mean velocity is equal to $\frac{1}{\sqrt{2}}$ times its maximum velocity.

Explanation: (a) let $x = a \sin \omega t$ be a periodic SHM

Let m be the mass-executing SHM

$v = \frac{dx}{dt}$

= $a\omega \cos \omega t$

$v_{max} = \alpha \omega ..... (since \cos \omega t = 1)$

Now, $\frac{K.E.}{P.E}$ = Total mechanical energy

Thus, $T.E. = \frac{1}{2} ma^{2} \omega ^{2}$

Thus, $K.E. = \frac{1}{2} ma^{2} \omega ^{2}$

Or we can say that the average total energy is K.E.max

(b) Let, amplitude = a

Angular frequency = $\omega$

Thus, maximum velocity = αω, which varies according to the sine law.

Thus, the RMS value of a complete cycle = $-\frac{1}{\sqrt{2}} \alpha \omega .$

Thus, the average

$K.E. = \frac{1}{2} mv_{rms}^{2}$

= $\frac{1}{2}m \frac{1}{2} a^{2}\omega ^{2}$

= $\frac{1}{2}\left [m \frac{1}{2} a^{2}\omega ^{2} \right ]$

= $\frac{1}{2}\left [ \frac{1}{2} m _{max}^{2} \right ]$

= $\frac{1}{2} K.E. _{max}$

(c) $v = a\omega \cos \omega t$

$V_{mean} = \frac{(a\omega -a\omega )}{2}$

Vmean = 0 ….. since vmax is not equal to vmean

$V_{rms} = \sqrt{\frac{v_{1}^{2}+v_{2}^{2}}{2} }= \frac{a\omega }{\sqrt{2}}$

Thus, $v_{rms }= \frac{v_{max}}{\sqrt{2}}$

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