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Question

A particle is executing SHM along a straight line. Its velocities at distances x1, and x2 from the mean position are v1 and v2 respectively. Its time period is

A
2πx22x21v21v22
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B
2πv21+v21x21+x22
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C
2πv21v21x21x22
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D
2πx21+x22v21+v22
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Solution

The correct option is B 2πx22x21v21v22
Let A be the amplitude of oscillation, then,

v21=ω2(A2x21) .....(1)

v22=ω2(A2x22) ....{2}

Subtracting eqn (1) from eqn (2), we get

v21v22=ω2(x22x21)

ω= v21v22x22x21

2πT= v21v22x22x21

Time Period, T=2π x22x21v21v22

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