Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A mass attached to a spring is free to oscillate with angular velocity ω, in a horizontal plane without friction or damping.

Q. 14.25 A mass attached to a spring is free to oscillate, with angular velocity \omega, in a horizontal plane without friction or damping. It is pulled to distance  x_{0} and pushed towards the centre with a velocity v_{0} at time t=0 Determine the amplitude of the resulting oscillations in terms of the parameters \omega, x_{0} and v_{0} . [Hint : Start with the equation x=a\; cos(\omega t+\theta ) and note that the initial velocity is negative.]

Answers (1)

best_answer

At the maximum extension of spring, the entire energy of the system would be stored as the potential energy of the spring.

Let the amplitude be A

\\\frac{1}{2}kA^{2}=\frac{1}{2}mv_{0}^{2}+\frac{1}{2}kx_{0}^{2}\\ A=\sqrt{x_{0}^{2}+\frac{m}{k}v_{0}^{2}}

The angular frequency of a spring-mass system is always equal to \sqrt{\frac{k}{m}}

Therefore

A=\sqrt{x_{0}^{2}+\frac{v_{0}^{2}}{\omega ^{2}}}

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question