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Home / Doubts and Solutions/JEE/Physics

A simple pendulum is suspended from a peg on a vertical wall.The pendulum is pulled away from the wall to a horizontal position and released.The ball hits the wall ,the coefficient of restitution being 2/(5^1/2) .What is the minimum no. Of collision after which the amplitude of oscillations becomes less than 60 degrees?

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Asked by m.nilu 21st September 2018, 6:05 PM
Answered by Expert
Answer:
As shown in figure, when the ball makes the first impact on the wall, let its velocity be u.
 
since potential energy is fully converted to kinetic energy, we have     (1/2)×m×u2 = m×g×L  or    u = begin mathsize 12px style square root of 2 g L end root end style .................(1)
after the first imapact, the ball will return with velocity  e×u, where e is the coefficient of restitution.
Since velocity after impact against the wall is less than the velocity before impact,
maximum height reached is keep on decreasing for every oscillation.
 
Let for nth oscillation, maximum height reached to a height h so that simple pendulam makes an angle 60º with vertical wall. In this case the height reached by the ball is L/2. Also for nth oscillation, ball will start from the wall with velocity en×u
 
hence  begin mathsize 12px style e to the power of n cross times u space equals space e to the power of n cross times square root of 2 cross times g cross times L end root space space equals space space square root of 2 cross times g cross times L over 2 end root space equals space square root of g cross times L end root end style...................(20
( note that e is coefficient of restitution, not the exponent-e)
by substituting the given coefficient of restitution, we get from (2),  begin mathsize 12px style open parentheses fraction numerator 2 over denominator square root of 5 end fraction close parentheses to the power of n space equals space fraction numerator 1 over denominator square root of 2 end fraction space space comma space b y space t a k i n g space log r i t h m space o n space b o t h space s i d e comma space space n open curly brackets log square root of 5 space minus space log space 2 close curly brackets space equals space log space 2 end style
or n = 6.2125 .  Hence after seventh collision, amplitude of oscillation will be less than 60º
Answered by Expert 22nd September 2018, 1:11 PM
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