Question 278: Oscillations

Question

A particle performing S.H.M. with maximum velocity ' $V$ '. If the amplitude double and periodic time is made, $\left(\frac{1}{3}ight)^{	ext {rd }}$ then the maximum velocity is

Accepted Answer

In simple harmonic motion (S.H.M.), the maximum velocity ($v_{	ext{max}}$) of a particle can be given by the formula:
$$ v_{	ext{max}} = A \omega $$
where $ A $ is the amplitude and $ \omega $ is the angular frequency. The angular frequency $ \omega $ can be related to the period $ T $ by:
$$ \omega = \frac{2\pi}{T} $$
Initially, the particle has an amplitude $ A $ and period $ T $, so the initial maximum velocity is:
$$ V = A \cdot \frac{2\pi}{T} $$
Now, the amplitude is doubled to $ 2A $ and the period becomes $ \left( \frac{1}{3} ight)T $. The new angular frequency $ \omega' $ is:
$$ \omega' = \frac{2\pi}{(1/3)T} = \frac{6\pi}{T} $$
Thus, the new maximum velocity $ v'_{	ext{max}} $ is:
$$ v'_{	ext{max}} = (2A) \cdot \frac{6\pi}{T} = 12A \cdot \frac{\pi}{T} $$
Since the original maximum velocity $ V $ was:
$$ V = A \cdot \frac{2\pi}{T} $$
Relating the new maximum velocity to the original, we get:
$$ v'_{	ext{max}} = 6 \cdot (A \cdot \frac{2\pi}{T}) = 6V $$
Therefore, the maximum velocity, when the amplitude is doubled and the period is reduced to one-third, is:
Option A: 6V

Answer

6V

Answer

$\frac{\mathrm{V}}{3}$

Answer

$\frac{3 V}{2}$

Answer

$\frac{2 \mathrm{~V}}{3}$

Oscillations

A particle performing S.H.M. with maximum velocity ' V '. If the amplitude double and periodic time is made, \left(\frac{1}{3}\right)^{\text {rd }} then the maximum velocity is

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