Q.

A body is performing SHM. Then its

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a

all are true

b

root mean square velocity is \large \frac{1}{{\sqrt 2 }} times of its maximum velocity

c

average total energy per cycle is equal to its maximum kinetic energy

d

average kinetic energy per cycle is equal to half of its maximum kinetic energy

answer is D.

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Detailed Solution

Suppose equation of SHM is \large x\, = \,A\cos \omega t

\large \therefore \,v\, = \,\frac{{dx}}{{dt}}\, = \, - A\omega \sin \omega t

\large \therefore \,{v_{\max }}\, = \,A\omega

\large \int\limits_0^T {{v^2}dt} \, = \,\int\limits_0^{\frac{{2\pi }}{\omega }} {{A^2}{\omega ^2}{{\sin }^2}\omega tdt\, = \,\pi }

\large \therefore \;{V_{rms}}\, = \,\sqrt {\frac{{\int\limits_0^T {{v^2}dt} }}{T}} \,\, = \,\frac{{A\omega }}{{\sqrt 2 }}\, = \,\frac{{{V_{\max }}}}{{\sqrt 2 }}

\large {\left( {KE} \right)_{\max }}\, = \,\frac{1}{2}mv_{\max }^2\, = \,\frac{1}{2}m{A^2}{\omega ^2}

\large PE\, = \,\frac{1}{2}m{\omega ^2}{x^2} = \,\frac{1}{2}m{\omega ^2}{A^2}co{s^2}\omega t

\large KE\, = \,\frac{1}{2}m{v^2}

\large KE\, = \,\frac{1}{2}m{A^2}{\omega ^2}{\sin ^2}\omega t

\large \therefore Total energy = KE+PE = \large \frac{1}{2}m{v^2}{A^2}

so option - (4) is right.

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