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Maximum speed of a particle in simple harmonic motion is vmax.The average speed of particle in SHM is equal to
A. vmax2
B. vmaxπ
C. πvmax2
D. 2vmaxπ

Answer
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Hint:We know that the general equation for a particle performing SHM is x=Asin(ωt+ϕ). On differentiating we get the equation for velocity v(t)=|Aωcos(ωt+ϕ)|. To get the average speed of the particle we integrate this equation for one time period, avg speed =1T0Tv(t)dt.

Complete step by step answer:
We know that the general equation for a particle performing simple harmonic motion is given as x=Asin(ωt+ϕ).
Where x is the displacement, A is the amplitude, ω is the angular frequency, ϕ is the initial phase and t is time.

Differentiating this equation with respect to time to get the equation for velocity v(t)=|Aωcos(ωt+ϕ)|.
To get the average speed of the particle we use the following equation,
avg speed=1T0Tv(t)dt.
Substituting the values we get,
avg speed=1T0TAωcos(ωt+ϕ)
We know that time period T can be written as 2πω.

Therefore the equation becomes,
Avg speed =12π/ω02πωAωcos(ωt+ϕ)
Avg speed =2Aωπ
Max value of velocity is obtained when cos(ωt+ϕ)=1 and is equal to vmax=Aω

Therefore the average speed of particles in SHM is 2vmaxπ.

Note:For a particle performing SHM the acceleration of the body is directly proportional to the displacement of the body from the mean position. The general equation for displacement of a particle performing simple harmonic motion is x=Asin(ωt+ϕ). Differentiating this equation we get the equation for velocity with time and acceleration of the particle. Average speed is defined as the rate of change of distance with respect to time. It can be written as a ratio of total distance travelled to the total time taken.