A small ball is suspended by a thread of length l = 1 m at the point O on the wall, forming a small angle $α = 2^{\circ}$ with the vertical (as shown in figure). Then the thread with ball was deviated through a small angle $β = 4^{\circ}$ and set free. Assuming the collision of the ball against the wall to be perfectly elastic, the oscillation period of such a pendulum (in seconds) is ___________ (Take $g = π^{2}$ ).

Moderate

Solution

The time period of free, oscillation of pendulum
$T = 2 π \sqrt{\frac{l}{g}} = 2 s$
Time taken by bob to go from extreme position A to mean position B is $= \frac{T}{4}$
Time taken by bob to move from mean position B to position C (where its angular displacement $α$ is half the angular amplitude $β$ ) is found from equation $α = βsin (\frac{2 π}{T} t)$ .
Solving we get $t = \frac{T}{12}$
$\Rightarrow$ Total time period of oscillation
$= 2 [\frac{T}{4} + \frac{T}{12}] = \frac{2}{3} T = \frac{4}{3} s$

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