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

The Correct Option is A

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 (V_max) occurs when the cosine function is at its maximum value of 1, which happens at ωt = 0. 
v_max = Aω 
V_avg =\(\frac { total \ distance\  traveled}{time\  taken}\) 
The total distance traveled in one complete cycle is 2A, and the time taken for one complete cycle is T.
V_avg = \(\frac {2A}{T}\) 
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: 
V_avg =\(\frac {2A}{1/f}\) = 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: 
V_avg = \(\frac {2πfA}{2π}\) = fA 
Given that V_max = Aω, 
we can rewrite V_avg as: 
V_avg = \(\frac {(V_{max} / ω)ω}{2π}\) = \(\frac {V_{max}}{ω} \times \frac {ω}{2π}\)= \(\frac {2V_{max}}{π}\)
Therefore, the correct option is (D) \(\frac {2V}{π.}\)