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Question

A particle executes simple harmonic motion (amplitude = A) between  x=-A and x=+A . The time taken for it to go from 0 to A2 is T1 and to go from A2 to A is T2. Then 


A
T1<T2
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B
T1>T2
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C
T1=T2
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D
T1=2T2
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Solution

The correct option is A T1<T2

Using  x=A sinωt
For x=A2, sin ωT1=12 T1=π6ω
For x=A, sin ω(T1+T2) =1  T1+T2=π2ω
 T2=π2ωT1=π2ωπ6ω=π3ω.i.e.T1< T2

Alternate method: In S.H.M., velocity of particle also oscillates simple harmonically. Speed is more near the mean position and less near the extreme position. Therefore the time taken for the particle to go from 0 to A2 will be less than the time taken to go from A2  to A. Hence T1< T2


Physics

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