Physics - SHM Question with Solution | TestHub

When a spring is in equilibrium, its extension$=\frac{mg}{k}$At the highest point of block, spring is in its natural length.

Hence, amplitude,$A=\frac{mg}{k}$

Given,$f=10 \mathrm{H}\mathrm{z}$

$\omega =2\pi f=20\pi \mathrm{r}\mathrm{a}\mathrm{d} {\mathrm{s}}^{-1}$

$\therefore \omega =\sqrt{\frac{k}{m}}$

On squaring both sides, we get

$\frac{k}{m}={\omega }^2$

$\Rightarrow v_{\mathrm{m}\mathrm{a}\mathrm{x}}=A\omega$

$=\frac{m}{k}g\omega =\frac{g\omega }{{\omega }^2}$

$=\frac{g}{\omega }=\frac{{\pi }^2}{20\pi }=\frac{\pi }{20}\mathrm{ m} {\mathrm{s}}^{-1}$

$\therefore A=\frac{mg}{k}=\frac{g}{{\omega }^2}$

$=\frac{{\pi }^2}{400 {\pi }^2}=\frac{0.25}{100}\mathrm{ m}=0.25 \mathrm{ c}\mathrm{m}$

Now, when the extension is$0.20 \mathrm{ c}\mathrm{m},$then displacement from the equilibrium position$=0.25-0.20=0.05 \mathrm{ c}\mathrm{m}$

$x=0.05 \mathrm{ c}\mathrm{m}$

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

$=20\pi \sqrt{{\left(0.25\right)}^2-{\left(0.05\right)}^2}$

$=2\pi \sqrt{6} \mathrm{ c}\mathrm{m} {\mathrm{s}}^{-1}$

$=15.4\mathrm{ }\mathrm{c}\mathrm{m} {\mathrm{s}}^{-1}$