A particle is performing simple harmonic motion along x axis with amplitude 4 cm and time period 1.2 sec. The minimum time taken by the particle to move from x = + 2 cm to x = + 4 cm and back again is :
0.4 s
0.3 s
0.2 s
0.6 s
Two springs of constants k1 and k2 have equal maximum velocities when executing simple harmonic motion. The ratio of their amplitudes (masses are equal) will be:
A rectangular block of mass m and the area of cross-section A floats in a liquid of density .If it is given a small vertical displacement from equilibrium it undergoes oscillation with a time period T. Then:
A simple harmonic oscillator has an amplitude and time period T. The time required by it to travel from x = to x = is:
Two points are located at a distance of 10m and 15 m from the source of oscillation. The period of oscillation is 0.005 s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillations of two points?
The total energy of a body performing S.H.M. depends on :
k,a,m
k,a,x
k,a
k,x
The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of oscillations is:
Which one of the following is a simple harmonic motion ?
Ball bouncing between two rigid vertical walls
Particle moving in a circle with uniform speed
Wave moving through a string fixed at both ends
Earth spinning about its own axis
The time period of a simple pendulum is 2 s. If its length is increased by 4 times, then its period becomes:
16 s
12 s
8 s
4 s
A spring has a certain mass suspended from it and its period of vertical oscillations is T. The spring is now cut into two halves and the same mass is suspended from one of the halves. Now the period of vertical oscillations is:
2 T
T/2
