Oscillations - Result Question 39
42. A particle is executing SHM along a straight line. Its velocities at distances $x_1$ and $x_2$ from the mean position are $V_1$ and $V_2$, respectively. Its time period is
[2015]
(a) $2 \pi \sqrt{\frac{x_2^{2}-x_1^{2}}{V_1^{2}-V_2^{2}}}$
(b) $2 \pi \sqrt{\frac{V_1^{2}+V_2^{2}}{x_1^{2}+x_2^{2}}}$
(c) $2 \pi \sqrt{\frac{V_1^{2}-V_2^{2}}{x_1^{2}-x_2^{2}}}$
(d) $2 \pi \sqrt{\frac{x_1^{2}-x_2^{2}}{V_1^{2}-V_2^{2}}}$
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Answer:
Correct Answer: 42. (a)
Solution:
- (a) As we know, for particle undergoing SHM,
$V=\omega \sqrt{A^{2}-X^{2}}$
$V_1^{2}=\omega^{2}(A^{2}-x_1^{2})$
$V_2^{2}=\omega^{2}(A^{2}-x_2^{2})$
Substracting we get,
$\frac{V_1^{2}}{\omega^{2}}+x_1^{2}=\frac{V_2^{2}}{\omega^{2}}+x_2^{2}$
$\Rightarrow \quad \frac{V_1^{2}-V_2^{2}}{\omega^{2}}=x_2^{2}-x_1^{2}$
$\Rightarrow \omega=\sqrt{\frac{V_1^{2}-V_2^{2}}{x_2^{2}-x_1^{2}}}$
$\Rightarrow T=2 \pi \sqrt{\frac{x_2^{2}-x_1^{2}}{V_1^{2}-V_2^{2}}}$
(c) As, we know, in SHM
Maximum acceleration of the particle, $\alpha=A \omega^{2}$ Maximum velocity, $\beta=A \omega$
$\Rightarrow \omega=\frac{\alpha}{\beta}$
$\Rightarrow T=\frac{2 \pi}{\omega}=\frac{2 \pi \beta}{\alpha} \quad[\because \omega=\frac{2 \pi}{T}]$
