Question

The maximum speed of a particle in S.H.M. is V. The average speed is

• A. $\frac {3V}{π}$
• B. $\frac {4V}{π}$
• C. $\frac {V}{π}$
• D. $\frac {2V}{π}$

Answer 1

Answer: (A)

$\frac {3V}{π}$

Answer 2

In SHM, the displacement of the particle
x(t) = A sin(ωt)
The velocity of the particle at any given time t:
v(t) = $\frac {dx}{dt}$ = Aω cos(ωt)
The maximum speed (Vmax) occurs when the cosine function is at its maximum value of 1, which happens at ωt = 0.
vmax = Aω
Vavg =$\frac { total \ distance\ traveled}{time\ taken}$
The total distance traveled in one complete cycle is 2A, and the time taken for one complete cycle is T.
Vavg = $\frac {2A}{T}$
The time period (T) can be calculated as the inverse of the frequency (f):
T = $\frac {1}{f}$
Substituting the expression for T, we get:
Vavg =$\frac {2A}{1/f}$ = 2Af = 2πfA
Since the angular frequency (ω) is related to the frequency (f) by ω = 2πf, we can rewrite the expression for V_avg as:
Vavg = $\frac {2πfA}{2π}$ = fA
Given that Vmax = Aω,
we can rewrite Vavg as:
Vavg = $\frac {(V_{max} / ω)ω}{2π}$ = $\frac {V_{max}}{ω} \times \frac {ω}{2π}$= $\frac {2V_{max}}{π}$
Therefore, the correct option is (D) $\frac {2V}{π.}$