TestHub
TestHub

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

Doubts & Discussion

Loading discussions...