Question 263: Oscillations

Question

A particle executing S.H.M. has velocities ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' at distances ' $x_1$ ' and ' $x_2$ ' respectively, from the mean position. Its frequency is

Accepted Answer

Particle velocities are
$$\begin{aligned}
& V_1^2=\omega^2\left(A^2-x_1^2ight) \quad	ext{.... (i)}\
& V_2^2=\omega^2\left(A^2-x_2^2ight) \quad	ext{.... (ii)}
\end{aligned}$$
Subtracting (i) from (ii),
$$\begin{aligned}
& V_2^2-V_1^2=\omega^2\left(x_1^2-x_2^2ight) \
& \omega=\sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}
\end{aligned}$$
As $\omega=2 \pi f$ we get,
$$\mathrm{f}=\frac{1}{2 \pi} \sqrt{\frac{\mathrm{~V}_2^2-\mathrm{V}_1^2}{\mathrm{x}_1^2-\mathrm{x}_2^2}}$$

Answer

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

Answer

$2 \pi \sqrt{\frac{x_1^2-x_2^2}{v_1^2-V_2^2}}$

Answer

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

Answer

$2 \pi \sqrt{\frac{\mathrm{x}_1^2-\mathrm{x}_2^2}{\mathrm{~V}_2^2-\mathrm{V}_1^2}}$

Oscillations

A particle executing S.H.M. has velocities ' \mathrm{V}_1 ' and ' \mathrm{V}_2 ' at distances ' x_1 ' and ' x_2 ' respectively, from the mean position. Its frequency is

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