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Oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsOscillations1 Mark
Question
A particle is vibrating in SHM when the displacements of the particle from its equilibrium position are $x_1$ and $x_2$, it has velocities $v_1$ and $v_2$ respectively. Show that its time period is given by $\text{T}=2\pi\sqrt{\frac{\text{x}_1^2-\text{x}_2^2}{\text{v}_2^2-\text{v}_1^2}}$.
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Answer

The particle velocity in SHM is given by: $\text{v}=\omega\sqrt{\text{A}^2-\text{x}^2}$ where A is the amplitude of oscillation. For displacement $\text{x}=\text{x}_1$$\text{v}_1=\omega\sqrt{\text{A}^2-\text{x}_1^2}$
$\text{v}_1^2=\omega^2(\text{A}^2-\text{x}_1^2)\cdots\text{(i)}$
For displacement $\text{x}=\text{x}_2$$\text{v}_2=\omega\sqrt{\text{A}^2-\text{x}_2^2}$
$\text{v}_2^2=\omega^2(\text{A}^2-\text{x}_2^2)\cdots\text{(ii)}$
Subtracting (i) from (ii), we have$\text{v}_2^2=\text{v}_2\omega^2(\text{x}_1^2-\text{x}_2^2)$
$\Rightarrow\omega=\sqrt{\frac{(\text{v}_2^2-\text{v}_1^2)}{(\text{x}_1^2-\text{x}_2^2)}}$
$\therefore$ Period of oscillation $\text{T}=\frac{2\pi}{\omega}$
$=2\pi\sqrt{\frac{(\text{x}_2^2-\text{x}_2^2)}{(\text{v}_1^2-\text{v}_2^2)}}$

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