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Question

Physics Question on Oscillations

In a simple pendulum the breaking strength of the string is double the weight of the bob. The bob is released from rest when the string is horizontal. The string breaks when it makes an angle θ\theta With the vertical

A

θ=cos1(13)\theta ={{\cos }^{-1}}\left( \frac{1}{3} \right)

B

θ=cos1(23)\theta ={{\cos }^{-1}}\left( \frac{2}{3} \right)

C

θ=60o\theta ={{60}^{o}}

D

θ=zero\theta =zero

Answer

θ=cos1(23)\theta ={{\cos }^{-1}}\left( \frac{2}{3} \right)

Explanation

Solution

By law of conservation of energy loss in PE = gain in KE \Rightarrow mgl(1cosθ)=12mv2mgl(1-\cos \theta )=\frac{1}{2}m{{v}^{2}} \Rightarrow mv2l=2mg(1cosθ)\frac{m{{v}^{2}}}{l}=2\,mg(1-\cos \theta ) At point of breaking mgcosθ=mv2lmg\cos \theta =\frac{m{{v}^{2}}}{l} \therefore mgcosθ=2mg(1cosθ)mg\cos \theta =2mg(1-\cos \theta ) 2mg=3mgcosθ2mg=3mg\cos \theta cosθ=23\cos \theta =\frac{2}{3} θ=cos1(23)\theta ={{\cos }^{-1}}\left( \frac{2}{3} \right)