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
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=ω√A20−y2
Displacement of the particle at any time t,
y=A0cosωt=A0cos2πTt
As the velocity is given by
v=ω√A20−y2
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