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
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a
2πx21+x22V21+V22
b
2πx22-x21V21-V22
c
2πV21+V22x21+x22
d
2πV21-V22x21-x22
answer is B.
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Detailed Solution
As we know, for particle undergoing SHM,V = ωA2-X2V21 = ω2(A2-x21) and V22 = ω2(A2-x22)On subtracting the relationsV21 -V22 = ω2(x22-x21)ω = V21-V22x22-x21T = 2πx22-x21V21-V22
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