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One end of a rod of length $\mathrm{L}$ is fixed to a point on the circumference of a wheel of radius $\mathrm{R}$. The other end is sliding freely along a straight channel passing through the centre $\mathrm{O}$ of the wheel as shown in the figure below. The wheel is rotating with a constant angular velocity $\omega$ about $\mathrm{O}$. Taking $\mathrm{T}=\frac{2 \pi}{\omega}$ the motion of the rod is
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Verified Answer
The correct answer is:
not simple harmonic but periodic with a period of $\mathrm{T}$
$\begin{array}{l}
\cos \theta=\frac{\mathrm{R}^{2}+\mathrm{x}^{2}-\mathrm{L}^{2}}{2 \mathrm{Rx}} \\
\Rightarrow \mathrm{x}^{2}=2 \mathrm{Rx} \operatorname{cosec} \theta+\mathrm{L}^{2}-\mathrm{R}^{2}
\end{array}$
displacement of S.H. M. is in the form of $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}+\mathrm{c}$
Therefore it is not S. H. M.
It is S. H. M. only which.
It $\mathrm{L}=\mathrm{R}$.
$\begin{array}{l}
\Rightarrow \mathrm{x}^{2}=2 \mathrm{Rx} \cos \theta \\
\Rightarrow \mathrm{x}=2 \mathrm{R} \cos \theta \\
\mathrm{x}=2 \mathrm{R} \cos \omega \mathrm{t}[\text { S.H. M.] }
\end{array}$
But period of this motion is $\mathrm{T}$
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