Question

A particle is executing shm withh amplitude a and has maximum velocity $v_0$ .its speed at displacement $3a/4$ will be?

• A. $\frac{v_0\sqrt{7}}{4}$
• B. $\frac{v_0\sqrt{4}}{4}$
• C. $\frac{v_0\sqrt{7}}{3}$
• D. $\frac{v_0\sqrt{9}}{3}$

Answer 1

The correct option is A $\frac{v_0\sqrt{7}}{4}$

$\begin{array}{c}{V}_{max}=V_0=A\omega \\ V=\omega \sqrt{A^2-x^2}\\ =\omega \sqrt{A^2-{\left(\frac{3A}{4}\right)}^2}\\ =\omega \sqrt{A^2-\frac{9A^2}{16}}\\ =\omega \sqrt{\frac{16A^2-9A^2}{16}}\\ =\omega \sqrt{\frac{7A^2}{16}}\\ =\frac{\omega A}{4}\sqrt{7}\\ =\frac{v_0\sqrt{7}}{4}\end{array}$

Hence, Option $A$ is correct .