Verified Solution • 2025–26 Session moderate

QJEEM23Phdad7: JEE Main 2023 Physics - Class 11 JEEM23Phdad7 Solution: A particle of mass 250 g executes a simple harmonic motion…

A particle of mass 250 g executes a simple harmonic motion under a periodic force F=(−25 x)N\mathrm{F}=(-25~x)\mathrm{N} . The particle attains a maximum speed of 4 m/s during its oscillation. The amplitude of the motion is ___________ cm.

Step-by-Step Verified Solution

F=−25xF = - 25x

.250d2xdt2=−25x.250{{{d^2}x} \over {d{t^2}}} = - 25x

d2xdt2=−100x{{{d^2}x} \over {d{t^2}}} = - 100x

⇒ω=10\Rightarrow \omega = 10 rad/sec

& ωA=vmax⁡\omega A = {v_{\max }}

10 A=410\,A = 4

⇒A=0.4\Rightarrow A = 0.4 m

=40= 40 cm

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