Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • In Exercise 14.9, let us take the position of mass when the spring is unstreched as x = 0, (c) at the maximum compressed position.

Q. 14.10 (c) In Exercise 14.9, let us take the position of mass when the spring is unstretched as x=0,and the direction from left to right as the positive direction of x-axis. Give x as a function of time t for the oscillating mass if at the moment we start the stopwatch (t=0), the mass is

(c) at the maximum compressed position.

In what way do these functions for SHM differ from each other, in frequency, in amplitude or the initial phase?

Answers (1)

best_answer

Amplitude is A = 0.02 m

Time period is \\\omega

\\\omega =\sqrt{\frac{k}{m}}\\ \omega =\sqrt{\frac{1200}{3}}\\ \omega =20\ rad/s

(c) At t = 0 the mass is at the maximum compressed position.

x(0) = -A

\phi =\frac{3\pi }{2}

\\x(t)=0.02sin\left ( 20t + \frac{3\pi }{2}\right )\\ x(t)=-0.02cos(20t)

Here x is in metres and t is in seconds.

The above functions differ only in the initial phase and not in amplitude or frequency.

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

Board Exam Result 2025 Live: UK Board 10th, 12th results tomorrow; AP SSC, TS Inter, UP, CBSE updates
anam-khan

When will CBSE Class 12 results 2025 be declared? Last two years' trends
anam-khan

CBSE Class 10, 12 board exams 2025 from April 7 to 11 for sports students

Latest Question