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<T2Method 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 t2−t1=π6ω=T1 Time taken for a/2 to a will be t3−t2=π2ω−π6ω=2π6ω=π3ω=T2

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