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 of oscillation is

A
2π x22x21v21v22
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B
2π v21+v22x21+x22
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C
2π v21v22x21x22
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D
2π x21+x22v21+v22
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Solution

## The correct option is A 2π ⎷x22−x21v21−v22For SHM, the velocity of particle at a distance x from mean position is given by, v=ω√A2−x2 ⇒v1=ω√A2−x21 ⇒v21=ω2(A2−x21)......(i) Similarly, v22=ω2(A2−x22).....(ii) Subtracting (ii) from equation (i); v21−v22=ω2(x22−x21) ⇒ω= ⎷v21−v22x22−x21 So, T=2πω ⇒T=2π ⎷x22−x21v21−v22

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