The correct option is
A 3√3TALet us consider the situation shown in the phasor diagram.
When a particle is moving from
A to
B, let the change in phase of the particle be
θ=ωt.
Given, Time interval
t=T3
⇒θ=ωt=2πT×T3=2π3
Which means change in phase of the particle in moving from
A to
B is
120∘ on the reference circle.
From the phasor diagram, we can deduce that, the maximum average velocity for a time interval of
T3 would be from
P to
Q.
∴ Maximum average velocity
v=Displacement between P and QTime interval
⇒v=√3A2+√3A2(T/3)=3√3AT
Thus, option (a) is the correct answer.
Alternate solution:
Let us suppose that the particle starts at the mean position. The equation of SHM is given by
x=Asinωt ......(1)
Differentiating
(1) with respect to time, we get
v=Aωcosωt
Since the maximum value of speed is at
x=0, the maximum average velocity for a time interval
T3 would be from
−T6 to
T6.
Average velocity of particle in a time interval
t is given by
vav=∫t0vdt∫t0dt
⇒vav=3T∫T/6−T/6Aω cosωtdt
⇒vav=3AT(2sin(ωT6))
Since
ωT=2π,
⇒vav=3√3AT
Thus, option (a) is the correct answer.