Potential energy of a particle with mass m is U=k[x]3 , where k is a positive constant. The particle is oscillating about the origin on x-axis. If the amplitude of oscillation is a, then its time period, T is
The dimensions and unit of phase constant Φ is
If the reference particle P moves in uniform circular motion, its projection along a diameter of the circle executes
The equation of motion of a particle is x= 3cos(0.45t+π/4)m. Its maximum acceleration is
If the sign in equation F = -kx is changed what would happen to the motion of the oscillating body?
Choose the correct time period of the function sin ωt + cos ωt
The velocity and acceleration amplitudes of body executing simple harmonic motion is
At what distance from the mean position would the K.E of a particle in simple harmonic motion be equal to its potential energy?
Let say from some distance x, the KE = PE and as total energy must be conserved and TE = -½ kA2
Thus we get 2PE = ½ kA2
Thus we get 2kx2 = kA2
We get x = A / √2
What is the maximum Kinetic energy and minimum potential energy of a harmonic oscillator with amplitude 0.03m, force constant 4×105 N/m and total mechanical energy of 230 J.
A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is