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Questions  

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
2πV12+V22x12+x22
b
2πV12-V22x12-x22
c
2πx12+x22V12+V22
d
2πx22-x12V12-V22

detailed solution

Correct option is D

In SHM, velocities of a particle at distances x1 and x2 from mean position are given by V12=ω2a2-x12                                    . . . . .(I)V22=ω2a2-x22                                   . . . . .(II)From equations (i) and (ii), we get V12-V22=ω2x22-x12ω=V12-V22x22-x12  ∴  T=2πx22-x12V12-V22

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