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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 2π√x22−x21v21−v22Let A be the amplitude of oscillation, then,v21=ω2(A2−x21) .....(1)v22=ω2(A2−x22) ....{2}Subtracting eqn (1) from eqn (2), we getv21−v22=ω2(x22−x21)⟹ω= ⎷v21−v22x22−x21⟹2πT= ⎷v21−v22x22−x21Time Period, T=2π ⎷x22−x21v21−v22

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