A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is $10 \mathrm{~cm}/\mathrm{s}$. The distance of the particle from the mean position when its speed becomes $5 \mathrm{~cm}/\mathrm{s}$ is $\sqrt{\alpha} \mathrm{~cm}$, where $\alpha =$ ______.

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Elasticity & SHM

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, velocity of the particle is

10 cm / s

. The distance of the particle from the mean position when its speed becomes

5 cm / s

α cm

, where

α =

______.

Answer: 12

At the mean position

v_{ma x} = A ω

, so with

A = 4

cm,

v_{ma x} = 10

cm/s gives

ω = 2.5

rad/s.

At displacement

x

v = ω A^{2} - x^{2}

. Setting

v = 5

cm/s:

5 = 2.5 16 - x^{2} \Rightarrow 4 = 16 - x^{2} \Rightarrow x^{2} = 12

α = 12

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Integer answer