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# A sphere of radius r is kept on a concave mirror of radius of curvature R. The arrangement's horizontal table (the surface of the concave mirror is frictionless and sliding not rolling displaced from its equilibrium position and left, then it executes S.H.M. The period of oscillation will be: (A) $2 \pi \sqrt{\left[\dfrac{(\mathrm{R}-\mathrm{r}) 1.4}{\mathrm{g}}\right]}$(B) $2 \pi \sqrt{\left[\dfrac{(\mathrm{R}-\mathrm{r})}{\mathrm{g}}\right]}$(C) $2 \pi \sqrt{\left[\dfrac{\mathrm{Rr}}{\mathrm{g}}\right]}$(D) $2 \pi \sqrt{\left[\dfrac{\mathrm{R}}{\mathrm{gr}}\right]}$

Last updated date: 16th Jul 2024
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Hint: We know that angular acceleration is the change in angular velocity divided by time, while tangential acceleration is the change in linear velocity divided by time. People sometimes forget that angular acceleration does not change with radius, but tangential acceleration does. Acceleration is a change in velocity, either in its magnitude that is speed or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the speed might be constant.

We know that tangential acceleration, $a_{1}=-g \sin \theta=-g \theta$
$-\quad a_{1}=-g^{\prime} \dfrac{x}{(R-r)}$
Motion is SHM is with time period $T=2 \pi \sqrt{\dfrac{\text { displacement }}{\text { acceleration }}}=2 \pi \sqrt{\dfrac{x}{\dfrac{g x}{(R-r)}}}=2 \pi \sqrt{\dfrac{R-r}{g}}$