 # A particle is making simple harmonic motion along `
Question:

A particle is making simple harmonic motion along the $\mathrm{X}$-axis. If at a distances $\mathrm{x}_{1}$ and $\mathrm{x}_{2}$ 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 :

1. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}_{2}^{2}+\mathrm{x}_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$

2. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}_{2}^{2}+\mathrm{x}_{1}^{2}}{v_{1}^{2}+v_{2}^{2}}}$

3. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}}{v_{1}^{2}+v_{2}^{2}}}$

4. $\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{x}_{2}^{2}-\mathrm{x}_{1}^{2}}{v_{1}^{2}-v_{2}^{2}}}$

Correct Option: , 4

Solution:

$v^{2}=\omega^{2}\left(A^{2}-x^{2}\right)$

$\mathrm{A}^{2}=\mathrm{x}_{1}^{2}+\frac{\mathrm{v}_{1}^{2}}{\omega^{2}}=\mathrm{x}_{2}^{2}+\frac{\mathrm{v}_{2}^{2}}{\omega^{2}}$

$\omega^{2}=\frac{v_{2}^{2}-v_{1}^{2}}{x_{1}^{2}-x_{2}^{2}}$

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