The velocity of a particle in `S.H.M.` at positions `x_(1)` and `x_2` are `v_(1)` and `v_(2)` respectively. Determine value of time period and amplitu

Question

The velocity of a particle in `S.H.M.` at positions `x_(1)` and `x_2` are `v_(1)` and `v_(2)` respectively. Determine value of time period and amplitu

The velocity of a particle in `S.H.M.` at positions `x_(1)` and `x_2` are `v_(1)` and `v_(2)` respectively. Determine value of time period and amplitude.

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

`v = omegasqrt(A^(2)-x^(2)) rArr v^(2) = omega^(2)(A^(2) - x^(2))`
At position `x_(1), v_(1)^(2) = omega^(2)(A^(2) - x_(1)^(2))`……(i)
At position `x_(2) , v_(2)^(2) = omega^(2)(A^(2) - x_(2)^(2))` …..(ii)
Subtracting `(ii)` from `(i)` from `v_(1)^(2) - v_(1)^(2) = omega^(2)(x_(2)^(2) - x_(1)^(2)) rArr omega = sqrt((v_(1)^(2) - v_(2)^(2))/(x_(2)^(2) - x_(1)^(2)))`
Time period `t = (2pi)/(omega) rArr T = 2pisqrt((x_(2)^(2) - x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))`
Dividing `(i)` by `(ii) (v_(1)^(2))/(v_(2)^(2)) = (A^(2) - x_(1)^(2))/(A^(2) - x_(2)^(2)) rArr v_(1)^(2)A^(2) - v_(1)^(2) xx x_(2)^(2) = v_(2)^(2) A^(2) - v_(2)^(2)x_(1)^(2)`
So `A^(2) (v_(1)^(2) - v_(2)^(2)) = v_(1)^(2)x_(2)^(2) - v_(2)^(2)x_(1)^(2) rArr A = sqrt((v_(1)^(2)x_(2)^(2) - v_(2)^(2)x_(1)^(2))/(v_(1)^(2) - v_(2)^(2)))`