Physics - SHM Question with Solution | TestHub

Given $m=1\text{ kg}$, $k=4\text{ N/m}$, $v=20\text{ cm/s}=0.2\text{ m/s}$, and $A=0.4\text{ m}$.

The angular frequency $\omega$ is given by:
$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{4}{1}}=2\text{ rad/s}$

The velocity of a particle in S.H.M. is:
$v=\omega \sqrt{A^2-x^2}$

Substituting the given values:
$0.2=2\sqrt{(0.4{)}^2-x^2}$

$0.1=\sqrt{0.16-x^2}$

Squaring both sides:
$0.01=0.16-x^2$

$x^2=0.16-0.01=0.15$

$x=\sqrt{0.15}\text{ m}$

Answer: $\sqrt{0.15}\mathrm{m}$