Physics - SHM Question with Solution | TestHub

A pendulum has length of $0.4\mathrm{m}$ and maximum speed $4\mathrm{m}/\mathrm{s}$. When the length makes an angle $30^{\circ }$ with the horizontal, its speed will be $2\sqrt{3}\mathrm{m}/\mathrm{s}$. Explanation:
$h=L-L\cos 0=L(1-\cos \theta )$ $=0.4\left(1-\cos 60^{\circ }\right)$ $=0.4\left(1-\frac{1}{2}\right)$ $=0.4\times \frac{1}{2}$ $=0.2$ $\frac{1}{2}m{v}_1^2=\frac{1}{2}m{v}_2^2+mgh$ $v_1^2=v_2^2+2gh$ $\therefore v_2^2=v_1^2-2gh$ $=(4{)}^2-2\times 10\times 0.2$ $=16-4=12$ $\therefore v_2=\sqrt{12}=2\sqrt{3}m/s$