Physics - SHM Question with Solution | TestHub

Given amplitude $A=4\text{ cm}$.

Maximum velocity at the mean position is $v_{\text{max}}=A\omega =12\text{ cm/s}$.

$\omega =\frac{12}{4}=3\text{ rad/s}$.

The velocity at a distance $x$ from the mean position is given by $v=\omega \sqrt{A^2-x^2}$.

Substituting the given values:

$6=3\sqrt{4^2-x^2}$

$2=\sqrt{16-x^2}$

Squaring both sides:

$4=16-x^2$

$x^2=12$

$x=2\sqrt{3}\text{ cm}$.

Answer: $2\sqrt{3}\mathrm{cm}$