CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The velocity v and displacement x of a particle executing simple harmonic motion are related as

vdvdx=ω2x

At x=0,v=v0. Find the velocity v when the displacement becomes x.


A

v=v20ω2x2

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

v20+ω2x2

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

v=v0ω20x

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

None of these

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

The correct option is A

v=v20ω2x2


We have, vdvdx=ω2x

or, v dv=ω2x dx

or, vv0v dv=x0ω2x dx .......(i)

When summation is made on ω2 x dx the quantity to be varied is x. When summation is made on vdv the quantity to be varied is v. As x varies from 0 to x the velocity varies from v0 to v. Therefore, on the left the limits of integration are from v0 to v and on the right they are from 0 to x. Simplifying (i),

[12v2]vv0=ω2[x22]x0

or, 12(v2v20)=ω2x22

or, v2=v20ω2x2

or, v=v20ω2x2


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon