MyQuestionIcon

10

You visited us 10 times! Enjoying our articles? Unlock Full Access!

X-t to V-t Graph

A particle ex...

Question

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is:

A

2π/√3

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

B

1/ 2π√3

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

C

2π√3

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

D

√3/2π

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

Open in App

Solution

The correct option is A
2π/√3

Suggest Corrections

thumbs-up

20

Similar questions

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

2 c m

. When the particle is at

1 c m

from the mean position, the magnitudes of the velocity and the acceleration are equal. Then its time period (in seconds) is ________________.

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

3 c m

. When the particle is at

2 c m

from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

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

5 cm

. When the particle is at

4 cm

from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is

Q. A particle executes SHM with an amplitude of

2 c m

. When the particle is at

1 c m

from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Aftermath of SHM

PHYSICS

Watch in App

explore_more_icon

Explore more

X-t to V-t Graph

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon