Question

A particle executes simple harmonic motion with an amplitude of A cm. At what displacement from mean does it speed equal to $\frac{1}{3}rd$ of maximum speed?

• A. 0.5A cm
• B. 0.6A cm
• C. 0.7A cm
• D. 0.9A cm

Answer 1

The correct option is D 0.9A cm

$v=\omega \sqrt{A^2-x^2}$ and $V_{max}=A.\omega$

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

$\frac{A\text{/}\omega }{3}=\text{/}\omega \sqrt{A^2-x^2}$

$\frac{A^2}{9}=A^2-x^2$

$x^2=A^2-\frac{A^2}{9}=\frac{8A^2}{9}$

$x=\frac{2\sqrt{2}A}{3}=\frac{2.8A}{3}=0.9A$ cm