(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

a wall is inclined to an angle "a" to the horizontal plane. a pendulum of length "l" attached to the wall making an angle "b" with the horizontal (a >b)is let to oscillate. Collision between the bob and the wall is perfectly elastic.show that the time period of the oscillation is 2.sqrt{l/g}.{pi/2 + sin[-1](b/a)}.

2. Relevant equations

In solving this problem i have found that all other terms are same for a pendulum in normal condition except in place of 2.pi there is 2.{pi/2 + sin[-1](b/a)}.So my question is what is the basis that we generally take the angular span of s.h.m to be 2 pi? Can there be any exception ?

3. The attempt at a solution

I tried to solve the problem following way.

from it's hanging point to the mean position it goes an angular distance pi/2.

from mean position to the angle a it goes an angular distance sin[-1](b/a), i can prove it as following.

if the angular distance is theta then

[l.sin a].sin theta = l.sin b

taking a,b small,

sin theta = l.b/l.a

hence, theta = sin[-1](b/a)

So effective angular span =2.{pi/2 + sin[-1](b/a)}

so time period = 2.sqrt{l/g}.{pi/2 + sin[-1](b/a)}.

Please tell me whether I am right in my concept or not.

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# Homework Help: Is the angular span of a body executing s.h.m always 2.pi?

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