Question

A particle is performing harmonic motion if its velocity are `v_(1)` and `v_(2)` at the displecement from the mean position are `y_(1)` and `y_(2)` respectively then its time period is
A. ` 2pi sqrt((y_(1)^(2) + y_(2)^(2))/(v_(1)^(2) + v_(2)^(2)))`
B. ` 2pi sqrt((v_(2)^(2) - v_(2)^(2))/(y_(1)^(2) - y_(2)^(2)))`
C. ` 2pi sqrt((y_(1)^(2) - y_(2)^(2))/(v_(2)^(2) + v_(1)^(2)))`
D. ` 2pi sqrt((v_(1)^(2) + v_(2)^(2))/(y_(1)^(2) + y_(2)^(2)))`

Answer 1

Correct Answer - C
Velocity of a particle is `SHM v = omega sqrt(A^(2) - y^(2))`
`v_(1) = omega sqrt(A^(2) - y_(1)^(2)) rArr v_(1)^(2) = omega^(2) A^(2)- y_(1)^(2)`…..(i)
and `v_(2) = omega sqrt(A^(2) - y_(2)^(2)) rArr v_(2)^(2) = omega^(2) A^(2)- omega^(2)y_(2)^(2)`…..(ii)
from equations (i) and (ii) we get
`v_(1)^(2) -v_(2)^(2) = omega^(2)(y^_1)^(2) - y_(2)^(2))`
`omega = sqrt((v_(2)^(1) - v_(1)^(2))/(y_(1)^(2) - y_(2)^(2))) rArr T = (2pi)/(omega) = 2pi sqrt((y_(1)^(2) - y_(2)^(2))/(v_(2)^(2) - _(1)^(2))`