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$$=2$$∘ with the vertical (as$$ shown in $$figure). Then the thread with ball was deviated through a small angle β$$=4$$∘ 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).

Question

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$$=2$$∘ with the vertical (as$$ shown in $$figure). Then the thread with ball was deviated through a small angle β$$=4$$∘ 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).

Infinity Learn · Accepted Answer

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

$= 2 [\frac{T}{4} + \frac{T}{12}] = \frac{2}{3} T = \frac{4}{3} s$

Applications of SHM

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material