Time period of a back performing `SHM` is `2` second and it can travel to and fro from equilibrium position upto maximum `5cm`. At start the pendulum is at maximum displacement on right side of equilibrium position. Find displacement and time relation.
Text Solution
Verified by Experts
Displacement expression for `S.H.M. , x = A sin (omegat + phi)` Time period of `SHM's T = (2pi)/(omega) = 2s :. omega = rad//s` Amplitude of `SHM A = 5 cm , :. x = 5 sin(pit + phi)` Now at `t = 0`, displacement `x = 5 cm :. 5 = 5 sin(pi xx 0 + phi) rArr sin phi = 1 rArr phi = (pi)/(2)` Therefore, `x = 5 "sin" pi(t + 1/2) rArr x = 5 sin(pit + (pi)/(2)) rArr x = 5 cos pit`
The time-period of a simple pendulum is 2 s and it can go to and fro from equilibrium position at maximum distance of 5 cm. If at the start of the motion the pendulum is in the position of maximum displacement towards the right of the equilibrium position, then write the displacement equation of the pendulum.
Time period of a simple pendulum is 2s and it can go to and fro from equilibrium position at a maximum distance of 6cm. If at the start of the motion the pendulum is in the position of maximum displacement towards the right of the equilibrium position, then write the displacement equation of the pendulum.
The maximum displacement of the constituents of the medium from their equilibrium positions is called
The acceleration of a particle executing SHM at a distance of 3 cm from equilibrium position is 5 cm//s^(2) . Its acceleration at a distance of 2 cm from equilibrium position is
The time period of S.H.M. is 16 seconds and it starts motion from the equililbrium position, after two seconds the velocity is pi m//s . Then the displacement amplitude is
For a simple pendulum, the time period is 3 s and has a maximum displacement of 5 cm on either side of its mean position. If at the beginning of the motion, the pendulum is at maximum displacement from its mean position, towards its right, then derive the displacement equation of the pendulum.
The acceleration of a particle performing a S.H.M. is 0.12 m//s^(2) , when the particle is at a dis"tan"ce of 3 cm from its equilibrium position. What is the period of S.H.M. ?
A particle oscillates, according to the equation x = 5 cos (pit//2) m. Then the particle moves from equilibrium position to the position of maximum displacement in time of
