Q122 / 150 · MHT CET 2026 (18 April Shift 1) · Physics
A particle is performing simple harmonic motion about $x = 0$ with an amplitude '$a$' and periodic time T. The speed of the particle at $x = \dfrac{a}{3}$ will be
- A$\dfrac{2\pi a}{T}$
- B$\dfrac{4\pi a}{3T}$
- Correct answerC$\dfrac{4\sqrt{2}\,\pi a}{3T}$
- D$\dfrac{\sqrt{3}\,\pi^2 a}{2T}$
Solution
The velocity of a particle in simple harmonic motion at a distance $x$ from the mean position is given by
$v = \omega \sqrt{a^2 - x^2}$
where $\omega$ is the angular frequency and $a$ is the amplitude.
Given $x = \dfrac{a}{3}$ and $\omega = \dfrac{2\pi}{T}$, substituting these values gives
$v = \dfrac{2\pi}{T} \sqrt{a^2 - \left(\dfrac{a}{3}\right)^2}$
$v = \dfrac{2\pi}{T} \sqrt{a^2 - \dfrac{a^2}{9}}$
$v = \dfrac{2\pi}{T} \sqrt{\dfrac{8a^2}{9}}$
$v = \dfrac{2\pi}{T} \left(\dfrac{2\sqrt{2}a}{3}\right)$
$v = \dfrac{4\sqrt{2}\pi a}{3T}$
Answer: $\dfrac{4\sqrt{2}\,\pi a}{3T}$
$v = \omega \sqrt{a^2 - x^2}$
where $\omega$ is the angular frequency and $a$ is the amplitude.
Given $x = \dfrac{a}{3}$ and $\omega = \dfrac{2\pi}{T}$, substituting these values gives
$v = \dfrac{2\pi}{T} \sqrt{a^2 - \left(\dfrac{a}{3}\right)^2}$
$v = \dfrac{2\pi}{T} \sqrt{a^2 - \dfrac{a^2}{9}}$
$v = \dfrac{2\pi}{T} \sqrt{\dfrac{8a^2}{9}}$
$v = \dfrac{2\pi}{T} \left(\dfrac{2\sqrt{2}a}{3}\right)$
$v = \dfrac{4\sqrt{2}\pi a}{3T}$
Answer: $\dfrac{4\sqrt{2}\,\pi a}{3T}$
Hence, the correct answer is option: 'C'.