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Grade 12th passMechanics

A pendulum of length 1m hangs from an inclined wall. Suppose that this pendulum is released at an initial angle of 10° and it bounces off the wall elastically when it reaches an angle of -5° as shown in the figure.Take g = pie^2m/sec2 m/s². The period of this pendulum is ( in second)

Aditya Pathak

8 Years agoGrade 12th pass

1 Answer

Eshan

Approved Tutor Answer8 Years ago

We need to find the time taken by pendulum in four intervals-

1. From angular displacement A to 0.
2. From angular displacement 0 to -A/2
3. From angular displacement -A/2 to 0.
4. From angular displacement 0 to A,

For the interval 1, time taken is 1/4th of the complete SHM motion = $T_1=\dfrac{1}{4}2\pi\sqrt{\dfrac{l}{g}}=\dfrac{\pi}{2}\sqrt{\dfrac{l}{g}}$
This is also equal to that in interval 4.
For interval 2, time taken is that from taking phase angle from $\dfrac{\pi}{6} to \dfrac{\pi}{2}$ which is $\dfrac{\pi}{3\omega}=\dfrac{\pi}{3}\sqrt{\dfrac{l}{g}}=T_2=T_3$

Hence the total time period of the pendulum is $\pi \sqrt{\dfrac{l}{g}}(1+\dfrac{2}{3}) =\dfrac{5\pi}{3}\sqrt{\dfrac{l}{g}}$ .

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