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Question

A particle executes simple harmonic motion between x = -A and x = +A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 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
Method 1: Qualitative. The velocity of a body executing S.H. M is maximum at its centre and decreases as the body proceeds to the extremes. Therefore if the time taken for the body to go from O to A/2 is T1 and then to go A is T2. then obviously T1 < T2. .
Method 2: Quantitative. Any S.H.M. is given by the equation y =a sin ω t where y is the displacement of the body at any instant t. a is the amplitude and ωis the angular frequency. 
When x =0, ωt1 = 0    t1 = 0

When x = a/2, ωt2 = π/6 t2=π6ω
When x = a, ωt3 = π/2 t3=π2ω

Time taken for O to a/2 will be t2t1=π6ω=T1

Time taken for a/2 to a will be 

t3t2=π2ωπ6ω=2π6ω=π3ω=T2

Physics

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