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Question
A small block oscillates back and forth on a smooth concave surface of radius R ib Figure . Find the time period of small oscillation.
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Solution
The maximum tension in the string of an oscillating pendulum is double of the minimum tension. Find the angular amplitude.
Let the speed of bob of the pendulum at an angle \[\theta\] be v.
Using the principle of conservation of energy between the mean and extreme positions, we get:
\[\frac{1}{2}\]mv2 − 0 = mgl(1 − cos θ)
v2 = 2gl(1 − cos θ) ...(1)
In a moving pendulum, the tension is maximum at the mean position, whereas it is minimum at the extreme position.
Maximum tension at the mean position is given by
Tmax = mg + 2mg(1 − cos θ)
Minimum tension at the extreme position is given by
Tmin = m g cosθ
According to the question,
Tmax = 2Tmin
⇒ mg + 2mg − 2m g cosθ = 2m g cosθ
⇒ 3mg = 4 mg cosθ
\[\Rightarrow \cos \theta = \frac{3}{4}\]
\[ \Rightarrow \theta = \cos^{- 1} \frac{3}{4}\]
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