Physics - SHM Question with Solution | TestHub

The velocity of a particle executing S.H.M. at a distance $x$ from the mean position is given by:

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

Squaring both sides, we get:

$V^2={\omega }^2(A^2-x^2)$

For the two given positions, we can write:

$V_1^2={\omega }^2(A^2-x_1^2)$

$V_2^2={\omega }^2(A^2-x_2^2)$

Subtracting the first equation from the second gives:

$V_2^2-V_1^2={\omega }^2(A^2-x_2^2)-{\omega }^2(A^2-x_1^2)$

$V_2^2-V_1^2={\omega }^2(x_1^2-x_2^2)$

${\omega }^2=\frac{V_2^2-V_1^2}{x_1^2-x_2^2}$

$\omega =\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}$

Since the angular frequency $\omega =2\pi f$, the frequency $f$ is:

$f=\frac{1}{2\pi }\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}$

Answer: $\frac{1}{2\pi }\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}$