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 :
Correct Option: , 4
Solution:
$v^{2}=\omega^{2}\left(A^{2}-x^{2}\right)$
$\mathrm{A}^{2}=\mathrm{x}_{1}^{2}+\frac{\mathrm{v}_{1}^{2}}{\omega^{2}}=\mathrm{x}_{2}^{2}+\frac{\mathrm{v}_{2}^{2}}{\omega^{2}}$
$\omega^{2}=\frac{v_{2}^{2}-v_{1}^{2}}{x_{1}^{2}-x_{2}^{2}}$
$T=2 \pi \sqrt{\frac{x_{1}^{2}-x_{2}^{2}}{v_{2}^{2}-v_{1}^{2}}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.