A system is oscillating with undamped simple harmonic motion. The… - Vidyadip

MCQ

A system is oscillating with undamped simple harmonic motion. Then the

A
average total energy per cycle of the motion is its maximum kinetic energy.
B
average total energy per cycle of the motion is $\frac{1}{\sqrt{2}}$ times its maximum kinetic energy.
C
root mean square velocity is $\frac{1}{\sqrt{2}}$ times its maximum velocity
✓
$(A)$ and $(C)$ both

✓

Answer

Correct option: D.

$(A)$ and $(C)$ both

d
Total energy of system is $P E+K E$ remains constant, when system is at mean position, $x=0 \Rightarrow P E=0,$ and $K E=$ max. so average of total energy is equal to total energy at mean position which is maximum kinetic energy.

similarly, at extreme points, $K E=0$ and $P E=$ $max.$

so average of total energy is equal to total energy at extreme position which is maximum potential energy.

Now, if $x=A \sin \omega t \Rightarrow v=A \omega \cos \omega t$

Maximum velocity is $v_{\max }=A \omega$ and $\mathrm{rms}$ value of velocity is $A \omega / \sqrt{2}=v_{\max } / \sqrt{2}(\because \mathrm{rms}$ value of $\cos \theta$ is $1 / \sqrt{2}$ )

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