The instantaneous displacement of a simple pendulum oscillator is given by x=A cos (ωt+π4) . Its speed will be maximum at time
1. π4ω
2. π2ω
3. πω
4. 2πω
The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is
1 144π2m/sec2
2 144m/sec2
3. 144π2m/sec2
4. 288π2m/sec2
A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by
1. – A Kx
2. A cos (Kx)
3. A exp (– Kx)
4. A Kx
What is the maximum acceleration of the particle doing the SHM y=2sin[πt2+ϕ] where 2 is in cm
1. π2cm/s2
2. π22cm/s2
3. π4cm/s2
4. π4cm/s2
A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is
1. ±A
2. Zero
3. ±A2
4. ±A√2
The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)
1. U=KX22
2. U=KX2
3. U=K
4. U=KX
The angular velocity and the amplitude of a simple pendulum is ω and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is
1, X2ω2(a2-X2ω2)
2. X2/(a2-x2)
3. (a2-X2ω2)/X2ω2
4. (a2-x2)/X2
There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,F=-Kx where x is the displacement. The total energy of body depends upon -
1. K, x
2. K, a
3. K, a, x
4. K, a, v
The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)
1. 18E
2. 14E
3. 12E
4. 23E