One end of a rod of length L is fixed to a po | vedclass

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(C) Let the position of the point on the circumference be $(R \cos 	heta, R \sin 	heta)$,where $	heta = \omega t$. The other end of the rod is at a

Vedclass · Accepted Answer

not simple harmonic but periodic with a period of $T$

Vedclass · Answer

(C) Let the position of the point on the circumference be $(R \cos 	heta, R \sin 	heta)$,where $	heta = \omega t$. The other end of the rod is at a distance $x$ from the center $O$ along the horizontal axis. Using the Pythagorean theorem in the triangle formed by the rod,we have $L^2 = (x - R \cos 	heta)^2 + (R \sin 	heta)^2$. Expanding this,$L^2 = x^2 - 2xR \cos 	heta + R^2 \cos^2 	heta + R^2 \sin^2 	heta$,which simplifies to $L^2 = x^2 - 2xR \cos 	heta + R^2$. This is a quadratic equation in $x$: $x^2 - (2R \cos 	heta)x + (R^2 - L^2) = 0$. Solving for $x$,we get $x = R \cos 	heta + \sqrt{R^2 \cos^2 	heta - (R^2 - L^2)} = R \cos 	heta + \sqrt{L^2 - R^2 \sin^2 	heta}$. Since the motion depends on $\cos 	heta$ and $\sin^2 	heta$ (where $	heta = \omega t$),the position $x(t)$ is a periodic function with period $T = \frac{2 \pi}{\omega}$. However,because of the square root term,the motion is not simple harmonic (which requires $x(t) = A \cos(\omega t + \phi)$). Thus,the motion is periodic but not simple harmonic.

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