Question 280: Oscillations

Question

A particle is executing a linear simple harmonic motion. Let ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' are its speed at distance ' $x_1$ ' and ' $x_2$ ' from the equilibrium position. The amplitude of oscillation is

Accepted Answer

For S.H.M, velocity is given by,
$$\begin{aligned}
& \quad \mathrm{V}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \Rightarrow \mathrm{~V}^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}^2ight) \quad	ext{... (i)}\
	herefore & \quad \mathrm{V}_1^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}_1^2ight) \
& 	ext { and } \mathrm{V}_2^2=\omega^2\left(\mathrm{~A}^2-\mathrm{x}_2^2ight) \quad	ext{... [From (i)]}\
& \frac{\mathrm{V}_1^2}{\mathrm{~V}_2^2}=\frac{\omega^2\left(\mathrm{~A}^2-\mathrm{x}_1^2ight)}{\omega^2\left(\mathrm{~A}^2-\mathrm{x}_2^2ight)} \
& \mathrm{V}_1^2\left(\mathrm{~A}^2-\mathrm{x}_2^2ight)=\mathrm{V}_2^2\left(\mathrm{~A}^2-\mathrm{x}_1^2ight) \
& A=\sqrt{\frac{\mathrm{V}_1^2 \mathrm{x}_2^2-\mathrm{V}_2^2 \mathrm{x}_1^2}{\mathrm{~V}_1^2-\mathrm{V}_2^2}}
\end{aligned}$$

Answer

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

Answer

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

Answer

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

Answer

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

Oscillations

A particle is executing a linear simple harmonic motion. Let ' \mathrm{V}_1 ' and ' \mathrm{V}_2 ' are its speed at distance ' x_1 ' and ' x_2 ' from the equilibrium position. The amplitude of oscillation is

Choose the correct answer: