A particle executes SHM in a straight line. The maximum speed of the particle during its motion is $V_{m a x}$ . Then the average speed of the particle during the SHM is :

\frac{V_{m}}{π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{V_{m}}{2 π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2 V_{m}}{π}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3 V_{m}}{π}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{2 V_{m}}{π}$
$V m = A ω$
$V a v g = \frac{d i s t a n c e t r a v e l e d}{t i m e t a k e n}$
$= \frac{A}{T}$
$T = \frac{2 π}{ω}$
$V a v g = \frac{4 A \times ω}{2 π} = \frac{2 A \times ω}{π}$
$V a v g = \frac{2 V_{m}}{π}$

Suggest Corrections

Similar questions

Q. A particle executes SHM in a straight line. The maximum speed of the particle during its motion is

v_{0}

. Its average speed in one complete oscillation is

Q. A particle is executing SHM according to the equation

x = A cos ω t

. Average speed of the particle during the interval

0 \leq t \leq \frac{π}{6 ω}

is:

Q. A particle is executing SHM according to the equation,

x = A cos ω t

. Average speed of the particle during the interval

0 \leq t \leq \frac{π}{6 ω}

Q. A particle is executing SHM according to the equation,

x = A cos ω t

. Average speed of the particle during the interval

0 \leq t \leq \frac{π}{6 ω}

The equation of a particle executing SHM is $x = (5 m) s i n [(π s^{- 1}) t + \frac{π}{3}]$ . What is its amplitude, time period, maximum speed and velocity. At t = 1s?

Join BYJU'S Learning Program

A particle executes SHM in a straight line. The maximum speed of the particle during its motion is Vmax. Then the average speed of the particle during the SHM is :

The correct option is C 2VmπVm=AωVavg=distance traveledtime taken=ATT=2πωVavg=4A×ω2π=2A×ωπVavg=2Vmπ

A particle executes SHM in a straight line. The maximum speed of the particle during its motion is $V_{m a x}$ . Then the average speed of the particle during the SHM is :

The correct option is C $\frac{2 V_{m}}{π}$
$V m = A ω$
$V a v g = \frac{d i s t a n c e t r a v e l e d}{t i m e t a k e n}$
$= \frac{A}{T}$
$T = \frac{2 π}{ω}$
$V a v g = \frac{4 A \times ω}{2 π} = \frac{2 A \times ω}{π}$
$V a v g = \frac{2 V_{m}}{π}$