Question

A particle executing SHM has velocities $20$ cm/s and $16$ cm/s at displacements $4$ cm and $5$ cm from its mean position respectively. Its time period is

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

Answer 1

Answer: (A)

$\pi /2$ s

$v_1={\omega }_1\sqrt{A^2-x_1^2}$
$v_2={\omega }_1\sqrt{A^2-x_2^2}$
$\frac{v_1}{v_2}=\frac{\sqrt{A^2-x{{}_1}^2}}{\sqrt{A^2-x{{}_2}^2}};v{{}_1}^2{A}^2-v{{}_1}^{2}x{{}_2}^2=v{{}_2}^2{A}^2-v{{}_2}^{2}x{{}_1}^2$
$A^2=\frac{v{{}_1}^{2}x{{}_2}^2-v{{}_2}^{2}x{{}_1}^2}{v{{}_1}^2-v{{}_2}^2}=\frac{400\times 25-16\times 256}{400-256}$
$A^2=41$
Substituting $A^2=41$ we get ${\omega }^2=16$
$\omega =\frac{2\pi }{T}=4;T=\frac{2\pi }{4}=\frac{\pi }{2}sec$