0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4 cm. Its initial displacement is 2 cm towards positive extreme position. Its equation for displacement is
  1. x=0.04cos(24πt+π/6)
  2. x=0.04sin(24πt)
  3. x=0.04sin(24πt+π/6)
  4. x=0.04cos(24πt)

A
x=0.04sin(24πt)
B
x=0.04cos(24πt)
C
x=0.04sin(24πt+π/6)
D
x=0.04cos(24πt+π/6)
Solution
Verified by Toppr

Was this answer helpful?
80
Similar Questions
Q1
A particle performing SHM with frequency 10 Hz and amplitude 5 cm is initially in left extreme position. The equation of its displacement will be (x is in metre)
View Solution
Q2
A particle in SHM is described by the displacement function x(t)=acos(ωt+θ). If the initial (t=0) position of the particle is 1 cm and its initial velocity is π cm/s. The angular frequency of the particle is π rad/s, then its amplitude is x cm. Find x.

View Solution
Q3
9.The particle performing SHM with a frequency of 5Hzand amplitude 2cmis intiallyin the positive extreme position.The equation for its displacement is ------------
View Solution
Q4

A spring block system is doing SHM with Amplitude A and angular frequency ω.

At t=T, the particle is at extreme position. Write displacement versus time equation.


View Solution
Q5
A particle performs SHM with a frequency of 1/2 Hz. If the time is measured from its instantaneous rest position, then the new frequency (in Hz) when its displacement becomes equal to half of its amplitude is:
View Solution