Physics - SHM Question with Solution | TestHub

PhysicsSHMKinematicsEasy2 minPYQ_2023
PhysicsEasynumerical

A particle of mass250 gexecutes a simple harmonic motion under a periodic forceF=(–25x) N. The particle attains a maximum speed of4 m s-1during its oscillation. The amplitude of the motion is ______cm.

Answer:
40.00
Solution:

Given, force F=(–25x) N

⇒0.25a=-25x

⇒a=-100x

Now comparing it with standard equation, a=-ω2x we get

ω=10.

Now, vmax=ωA

⇒ωA=4

⇒A=410=0.4 m

⇒A=40 cm.

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Kinematics
⏱ 2mℹ️ Source: PYQ_2023

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