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(B) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by $v^{2} = \omega^{2}(A^{2} - x^{2})$,where $A$ is the am

Vedclass · Accepted Answer

$T=2 \pi \sqrt{\frac{x_{2}^{2}-x_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$

Vedclass · Answer

(B) The velocity $v$ of a particle in simple harmonic motion at a displacement $x$ is given by $v^{2} = \omega^{2}(A^{2} - x^{2})$,where $A$ is the amplitude and $\omega$ is the angular frequency.For the given positions:$v_{1}^{2} = \omega^{2}(A^{2} - x_{1}^{2}) \implies \frac{v_{1}^{2}}{\omega^{2}} = A^{2} - x_{1}^{2} \implies A^{2} = x_{1}^{2} + \frac{v_{1}^{2}}{\omega^{2}}$$v_{2}^{2} = \omega^{2}(A^{2} - x_{2}^{2}) \implies \frac{v_{2}^{2}}{\omega^{2}} = A^{2} - x_{2}^{2} \implies A^{2} = x_{2}^{2} + \frac{v_{2}^{2}}{\omega^{2}}$Equating the two expressions for $A^{2}$:$x_{1}^{2} + \frac{v_{1}^{2}}{\omega^{2}} = x_{2}^{2} + \frac{v_{2}^{2}}{\omega^{2}}$$\frac{v_{1}^{2} - v_{2}^{2}}{\omega^{2}} = x_{2}^{2} - x_{1}^{2}$$\omega^{2} = \frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}$Since $T = \frac{2\pi}{\omega}$,we have $\omega^{2} = \frac{4\pi^{2}}{T^{2}}$.$\frac{4\pi^{2}}{T^{2}} = \frac{v_{1}^{2} - v_{2}^{2}}{x_{2}^{2} - x_{1}^{2}}$$T^{2} = 4\pi^{2} \frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}$$T = 2\pi \sqrt{\frac{x_{2}^{2} - x_{1}^{2}}{v_{1}^{2} - v_{2}^{2}}}$

$A$ particle is making simple harmonic motion along the $x$-axis. If at a distance $x_{1}$ 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 -

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