MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Beat Frequency

A body execut...

Question

A body executes simple harmonic motion of amplitude 2 m. Its velocity while passing tgrough the mean position is 10m/s.Find its frequency

Open in App

Solution

At mean position Vmax=aω=a.2π/T=2×2π/T=4π/T But given that the velocity at mean position is 10m/s Therefore 10 = 4π/T Therefore ,frequency ,1/T =10/4π =.79

Suggest Corrections

thumbs-up

0

Similar questions

Q. A particle executes simple harmonic motion of amplitude

A

. At what distance from the mean position is its kinetic energy equal to its potential energy?

Q. The ratio between kinetic and potential energies of a body executing simple harmonic motion, when it is at a distance of

\frac{1}{N}

of its amplitude from the mean position is

Q. A particle of mass

m

executes simple harmonic motion with amplitude

a

v

. The average kinetic energy during its motion from the position of equilibrium to the end is

Q. If a body is executing simple harmonic motion and its current displacement is

\frac{\sqrt{3}}{2}

times the amplitude from its mean position, then the ratio between potential energy and kinetic energy is:

Q. A particle executes simple harmonic motion with an amplitude of

4 c m .

At the mean position the velocity of the particle is

10 c m / s

. The distance of the particle from the mean position when its speed becomes

5 c m / s

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Beats

PHYSICS

Watch in App

explore_more_icon

Explore more

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon