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Question

The displacement time graph of a particle executing (S.H.M.) is shown in figure. Whether the following statement is true or not?



A) The force is zero at t=3T4

B) The acceleration is maximum at t=4T4.

C) The velocity is maximum at t=T4.

D) The P.E. is equal to K.E of oscillation at t=T2


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Solution

A) Formula Used: F=ma

Let displacement of the particle at any time t,

y=A0cosωt=A0cos2πTt(ω=2πt)

So, the acceleration will be a=ω2y

a=ω2A0cos2πTt

Force F=mω2y.

At t=3T4, y=A0cos(2πT×3T4)

=A0cos(3π2)=0

Hence F=0.

Final Answer: True


B) Let displacement of the particle at any time t,

y=A0cosωt=A0cos2πTt

So, the acceleration will be A=ω2y

a=ω2A0cos2πTt.

Att=4T4, y=A0cos(2πT×T)

=A0cos2π=A0

|a|=ω2A0, which is maximum value of a.

Final Answer: True.


C) Formula Used: v=ωA20y2

Displacement of the particle at any time t,

y=A0cosωt=A0cos2πTt

As the velocity is given by

v=ωA20y2

At t=T4

y=A0cos(2πT×T4)=A0cos(π2)

y=0

Hence, v=ωA0, which is maximum value of velocity.

Final Answer: True.


D) Displacement of the particle at any time t,

y=A0cosωt=A0cos2πTt

At t=T2,

y=A0cos(2πT×T2)=A0

Displacement is maximum, i.e. corresponds to extreme position, it means PE is maximum and KE is zero.

Final Answer: Not true


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