JEE Main 2024 · Physics · Simple Harmonic Motion · easy

A particle executes simple harmonic motion along a straight line. When its displacement from the mean position is 3 cm its speed is 8 cm s^-1, and…

Numerical answer

A particle executes simple harmonic motion along a straight line. When its displacement from the mean position is

3\ \mathrm{cm}

its speed is

8\ \mathrm{cm\,s^{-1}}

, and when the displacement is

4\ \mathrm{cm}

its speed is

6\ \mathrm{cm\,s^{-1}}

. The maximum speed of the particle (in

\mathrm{cm\,s^{-1}}

) is

Show answer and solution

Answer: 10

Use

v^2=\omega^2(A^2-x^2)

twice:

64=\omega^2(A^2-9)

and

36=\omega^2(A^2-16)

. Subtracting,

28=7\omega^2

, so

\omega=2\ \mathrm{s^{-1}}

. Then

A^2-9=16

, giving

A=5\ \mathrm{cm}

. Maximum speed

v_{max}=\omega A=2\times5=10\ \mathrm{cm\,s^{-1}}

Original question written in the JEE Main 2024 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake