If the speed v of the bob in a simple pendulu | vedclass

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Question

(A) For a simple pendulum undergoing simple harmonic motion,the displacement $x$ is given by $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt}

Vedclass · Accepted Answer

$I$

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(A) For a simple pendulum undergoing simple harmonic motion,the displacement $x$ is given by $x = A \sin(\omega t)$.The velocity is $v = \frac{dx}{dt} = A\omega \cos(\omega t)$.The tangential acceleration is $a = \frac{dv}{dt} = -A\omega^2 \sin(\omega t)$.From these equations,we have:$\frac{v}{A\omega} = \cos(\omega t)$ and $\frac{a}{-A\omega^2} = \sin(\omega t)$.Using the identity $\sin^2(	heta) + \cos^2(	heta) = 1$,we get:$\left(\frac{v}{A\omega}ight)^2 + \left(\frac{a}{-A\omega^2}ight)^2 = 1$$\frac{v^2}{A^2\omega^2} + \frac{a^2}{A^2\omega^4} = 1$This is the equation of an ellipse in the $v-a$ plane,where $v$ is on the horizontal axis and $a$ is on the vertical axis. Thus,the correct graph is $I$.

If the speed $v$ of the bob in a simple pendulum is plotted against the tangential acceleration $a$,the correct graph will be represented by

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