Question 269: Oscillations

Question

A particle is performing S.H.M. about its mean position with an amplitude ' $a$ ' and periodic time ' $T$ '. The speed of the particle when its displacement from mean position is $\frac{a}{3}$ will be

Accepted Answer

In simple harmonic motion (S.H.M.), the displacement of a particle from the mean position is given by:
$$ x = a \cos(\omega t) $$
where:
$ a $ is the amplitude,
$ \omega $ is the angular frequency,
$ t $ is the time.
The angular frequency $\omega$ is related to the periodic time $T$ by:
$$ \omega = \frac{2\pi}{T} $$
The speed $v$ of a particle performing S.H.M. is given by:
$$ v = \omega \sqrt{a^2 - x^2} $$
Substituting the displacement $x = \frac{a}{3}$ into the equation, we get:
$$ v = \omega \sqrt{a^2 - \left(\frac{a}{3}\right)^2} $$
Calculate $a^2 - \left(\frac{a}{3}\right)^2$:
$$ a^2 - \left(\frac{a}{3}\right)^2 = a^2 - \frac{a^2}{9} = \frac{9a^2}{9} - \frac{a^2}{9} = \frac{8a^2}{9} $$
Therefore, the speed $v$ is:
$$ v = \omega \sqrt{\frac{8a^2}{9}} = \omega \cdot a \cdot \frac{2\sqrt{2}}{3} $$
Substitute $\omega = \frac{2\pi}{T}$:
$$ v = \frac{2\pi}{T} \cdot a \cdot \frac{2\sqrt{2}}{3} $$
Simplifying this expression gives:
$$ v = \frac{4 \sqrt{2} \pi a}{3 T} $$
Thus, the correct answer is Option B:
$$ \frac{4 \sqrt{2} \pi a}{3 T} $$

Answer

$\frac{2 \pi \mathrm{a}}{\mathrm{T}}$

Answer

$\frac{4 \sqrt{2} \pi \mathrm{a}}{3 \mathrm{~T}}$

Answer

$\frac{4 \pi^2 a}{3 \mathrm{~T}}$

Answer

$\frac{\sqrt{3} \pi^2 a}{2 \mathrm{~T}}$

Oscillations

A particle is performing S.H.M. about its mean position with an amplitude ' a ' and periodic time ' T '. The speed of the particle when its displacement from mean position is \frac{a}{3} will be

Choose the correct answer: