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(A) In simple harmonic motion,the velocity of the body is given by $v = \omega \sqrt{a^2 - x^2}$.The time $dt$ spent by the body in a small interval $

Vedclass · Accepted Answer

$x=\pm a$

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(A) In simple harmonic motion,the velocity of the body is given by $v = \omega \sqrt{a^2 - x^2}$.The time $dt$ spent by the body in a small interval $dx$ is given by $dt = \frac{dx}{v} = \frac{dx}{\omega \sqrt{a^2 - x^2}}$.The probability $P(x)dx$ of finding the body in the interval $dx$ is proportional to the time spent in that interval,i.e.,$P(x) \propto \frac{1}{v} = \frac{1}{\omega \sqrt{a^2 - x^2}}$.As $x 	o \pm a$,the velocity $v 	o 0$,which means the time spent $dt 	o \infty$.Therefore,the probability of finding the body is highest at the extreme positions $x = \pm a$.Thus,the correct option is $A$.

$A$ body is executing simple harmonic motion of amplitude $a$ and period $T$ about the equilibrium position $x=0$. Large numbers of snapshots are taken at random of this body in motion. The probability of the body being found in a very small interval $x$ to $x+|dx|$ is highest at

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