Question

A particle is executing SHM with amplitude $A$ and has maximum vekocity $v_0$. Find its speed when it is located at distance of $\frac{A}{2}$ from position.

Answer 1

we know,

$v_{max}$=$v_o$=A$\omega$

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

=$\omega$$\sqrt{A^2-(\frac{3A}{4}{)}^2}$

by solving this,

v=$\omega$A$\frac{\sqrt{7}}{4}$

=$v_o$$\frac{\sqrt{7}}{4}$