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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
2πV21+V22x21+x22
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B
2πV21V22x21x22
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C
2πx21+x22V21+V22
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D
2πx22x21V21V22
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Solution

The correct option is D 2πx22x21V21V22

Using the expressions for velocity for SHM,

v1=ωA2x21
v2=ωA2x22
Squaring both,
v21=ω2(A2x21) ....(3)
v22=ω2(A2x22) ....(4)
Subtracting (4) from (3), v21v22=ω2(x22x21)
4π2(x22x21)=T2(v21v22)
T=2π x22x21v21v22 where we have used ω=2πT


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