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PhysicsSHMEnergyHard2 minPYQ_2020
PhysicsHardsingle choice

The displacement time graph of a particle executing SHM is given in figure: (sketch is schematic and not to scale) 

Which of the following statements is/are true for this motion? 

(A) The force is zero at t=3T4
(B) The magnitude of acceleration is maximum at t=T
(C) The speed is maximum at t=T4
(D) The P.E. is equal to K.E. of the oscillation at t=T2

Question diagram: The displacement time graph of a particle executing SHM is g

Options:

Answer:
A
Solution:

From graph equation of SHM.

X=Acosωt
(A) at 3T4 particle at mean position
a=0
F=0
(B) at T particle again at extreme position so acceleration is maximum
(C) at t=T4, particle is at mean position so velocity maximum.   a=0
(D) KE=PE
12kA2-x2=12kx2
A2=2x2
x=+A2

A2=Acosωt

t=T8
 A, B and C are correct.

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy
2mℹ️ Source: PYQ_2020

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