Q.

Kinetic energy of a particle of mass 40 gm executing simple harmonic motion along a straight line is given by K=161−0.25x2  joule where x is in cm. Then frequency of oscillation of the particle is

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away
An Intiative by Sri Chaitanya

a

5002πHz

b

500πHz

c

250πHz

d

1000πHz

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

Kinetic energy K=12mω2A21−x2A2Comparing this equation with the given equation we get, A2=10.25⇒A=2 cmand 12mω2A2=16⇒12×401000.ω2.21002=16⇒ω2=16×50×2500 ω=10002 rad/s∴ f=ω2π=5002π Hz
Watch 3-min video & get full concept clarity
Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs
Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation
Result Producing online coaching for JEE/NEET of south India
Largest All India Test Series for JEE/NEET
Best Books for IIT JEE
Best Books for NEET
How to Join IIT after 10th
whats app icon
Kinetic energy of a particle of mass 40 gm executing simple harmonic motion along a straight line is given by K=161−0.25x2  joule where x is in cm. Then frequency of oscillation of the particle is