Question

The velocities of a particle performing linear SHM are $0.13$ m/s and $0.12$ m/s when it is at $0.12m$ and $0.13$ m from mean position respectively, the period is:

• A. $\pi$s
• B. 2$\pi$s
• C. 3$\pi$s
• D. 4$\pi$s

Answer 1

The correct option is B 2$\pi$s

$V=\omega \sqrt{A^2-x^2}$

$V_1=\omega \sqrt{A^2-x_1^2}$

$V_2=\omega \sqrt{A^2-x_2^2}$

$\frac{0.13}{0.12}$ = $\frac{\sqrt{A^2-(0.12{)}^2}}{\sqrt{A^2-(0.13{)}^2}}$

$\frac{{0.13}^2}{{0.12}^2}$ = $\frac{A^2-(0.12{)}^2}{A^2-(0.13{)}^2}$

Cross-multiplying,

$\left(\right(0.13{)}^2-\left(0.12{)}^2\right)A^2=(0.13{)}^4-(0.12{)}^4$

$A^2=(0.13{)}^2+(0.12{)}^2$

$A=\sqrt{(0.13{)}^2+(0.12{)}^2}$

Using, $x$ = $\omega \sqrt{A^2-x^2}$

$0.13=\omega \sqrt{(0.13{)}^2+(0.12{)}^2-(0.12{)}^2}$

$\therefore \omega =1$

$T=2\pi /\omega$

$T=2\pi \sec$