When a body is in S.H.M.,match the following: | vedclass

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(C) In $S.H.M.$,the displacement is $x = A \sin \omega t$.Velocity $v = \omega \sqrt{A^2 - x^2}$ and acceleration $a = -\omega^2 x$.$A$. Velocity is m

Vedclass · Accepted Answer

$A-II, B-III, C-IV, D-I$

Vedclass · Answer

(C) In $S.H.M.$,the displacement is $x = A \sin \omega t$.Velocity $v = \omega \sqrt{A^2 - x^2}$ and acceleration $a = -\omega^2 x$.$A$. Velocity is maximum at the mean position $(x = 0)$,so $A-II$.$B$. Kinetic Energy $(K.E.)$ is $\frac{1}{2} m \omega^2 (A^2 - x^2)$. For $K.E. = \frac{3}{4} E_{total} = \frac{3}{4} (\frac{1}{2} m \omega^2 A^2)$,we get $A^2 - x^2 = \frac{3}{4} A^2$,which implies $x^2 = \frac{1}{4} A^2$ or $x = \frac{A}{2}$. So $B-III$.$C$. Potential Energy $(P.E.)$ is $\frac{1}{2} m \omega^2 x^2$. For $P.E. = \frac{3}{4} E_{total} = \frac{3}{4} (\frac{1}{2} m \omega^2 A^2)$,we get $x^2 = \frac{3}{4} A^2$,which implies $x = \frac{\sqrt{3}}{2} A$. So $C-IV$.$D$. Acceleration is maximum at the extreme position $(x = A)$,so $D-I$.Thus,the correct match is $A-II, B-III, C-IV, D-I$.

List-$I$	List-$II$
$A$. Velocity is maximum	$I$. Acceleration is maximum
$B$. $K.E.$ is $\left(\frac{3}{4}\right)^{\text{th}}$ of total energy	$II$. At mean position
$C$. $P.E.$ is $\left(\frac{3}{4}\right)^{\text{th}}$ of total energy	$III$. At half of the amplitude
$D$. Acceleration is maximum	$IV$. At $\frac{\sqrt{3}}{2}$ times the amplitude

Similar Questions

$A$ horizontal platform with an object placed on it is executing $S.H.M.$ in the vertical direction. The amplitude of oscillation is $3.92 \times 10^{-3} \, m$. What must be the least period of these oscillations,so that the object is not detached from the platform (in $, s$)?

$A$ graph of the square of the velocity against the square of the acceleration of a given simple harmonic motion is

$Assertion :$ In simple harmonic motion,the velocity is maximum when the acceleration is minimum.
$Reason :$ Displacement and velocity of $S.H.M.$ differ in phase by $\frac{\pi }{2}$.

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