Physics - SHM Question with Solution | TestHub

The speed of a particle in simple harmonic motion at a distance $x$ from the equilibrium position is given by $V=\omega \sqrt{A^2-x^2}$, where $\omega$ is the angular frequency and $A$ is the amplitude.

Squaring both sides, we get:

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

For the two given positions, we have:

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

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

Dividing the two equations to eliminate ${\omega }^2$:

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

Cross-multiplying yields:

$V_1^2(A^2-x_2^2)=V_2^2(A^2-x_1^2)$

$V_1^2{A}^2-V_1^2{x}_2^2=V_2^2{A}^2-V_2^2{x}_1^2$

Rearranging the terms to solve for $A^2$:

$A^2(V_1^2-V_2^2)=V_1^2{x}_2^2-V_2^2{x}_1^2$

$A^2=\frac{V_1^2{x}_2^2-V_2^2{x}_1^2}{V_1^2-V_2^2}$

Taking the square root, the amplitude is:

$A=\sqrt{\frac{V_1^2{x}_2^2-V_2^2{x}_1^2}{V_1^2-V_2^2}}$

Answer: $\sqrt{\frac{V_1^2{x}_2^2-V_2^2{x}_1^2}{V_1^2-V_2^2}}$