JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Elasticity & SHM

A particle is making simple harmonic motion along the X-axis. If at a distances

x_{1}

and

x_{2}

from the mean position the velocities of the particle are

v_{1}

and

v_{2}

respectively. The time period of its oscillation is given as :

Answer: (A)

Solution

For SHM,

v^{2} = ω^{2} (A^{2} - x^{2})

, where

ω = \frac{2 π}{T}

and

A

is amplitude. At positions

x_{1}

and

x_{2}

v_{1}^{2} = ω^{2} (A^{2} - x_{1}^{2})

and

v_{2}^{2} = ω^{2} (A^{2} - x_{2}^{2})

. Subtracting:

v_{1}^{2} - v_{2}^{2} = ω^{2} (x_{2}^{2} - x_{1}^{2})

. Therefore,

ω^{2} = \frac{v _{1}^{2} - v _{2}^{2}}{x _{2}^{2} - x _{1}^{2}}

, giving

T = \frac{2 π}{ω} = 2 π \frac{x _{2}^{2} - x _{1}^{2}}{v _{1}^{2} - v _{2}^{2}}

JEE Main · 2021 · 20 Jul 2021 · Shift 2 · Single correct option