Question

Maximum speed of a particle in simple harmonic motion is vmax. Then average speed of a particle in shm is equal to

Answer 1

Dear Student,

◆ Answer -

vavg = 2vmax/π

● Explanation -

In S.H.M. maximum speed is given by -

vmax = Aω

Also in S.H.M. average speed is given by -

vavg = 2Aω/π

vavg = 2(Aω)/π

Put Aω = vmax,

vavg = 2vmax/π

Hence, average speed of a particle in S.H.M. is equal to 2vmax/π.

Thanks dear. Hope this helps you...

Answer 2

average speed of the particle in shm is 2Vmax/π

maximum speed of a particle in simple harmonic motion is Vmax.

we have to find average speed of particle in term of Vmax.

in simple harmonic motion,

y = Asinωt

so, v = ωA cosωt

then maximum speed of particle will be Vmax = ωA ........(1)

now average speed of particle, < v > = 1/π ∫ωAcosωt dt

= ωA/π [sinωt]^(π/2)_(-π/2)

= ωA/π [sin(π/2) - sin(-π/2) ]

= 2ωA/π

hence, average speed of the particle is <v> = 2Vmax/π = 0.636 Vmax

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