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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 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 x22x21v21v22
For SHM, the velocity of particle at a distance x from mean position is given by,
v=ωA2x2
v1=ωA2x21
v21=ω2(A2x21)......(i)
Similarly,
v22=ω2(A2x22).....(ii)
Subtracting (ii) from equation (i);
v21v22=ω2(x22x21)
ω= v21v22x22x21
So,
T=2πω
T=2π x22x21v21v22

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