If a simple pendulum is brought deep inside a mine from earth surface, its time period of oscillation will 

The amplitude of a simple harmonic oscillator is A and speed at the mean position is ${\mathrm{v}}_0$. The speed of the oscillator at the position $\mathrm{x} = \frac{\mathrm{A}}{\sqrt{3}}$ is:

The initial phase of the particle executing SHM with y = 4 sin $\mathrm{\omega }$t + 3 cos $\mathrm{\omega }$t is 

A spring-block system is brought from the earth's surface to deep inside mine. Its period of oscillation will:

A particle executes S.H.M with amplitude A. If the time taken by the particle to travel from -A to A/2 is 4 seconds, its time period is 

Two simple harmonic motions are represented by ${\mathrm{y}}_1 = 6 \sin \left(2\mathrm{\pi t} + \frac{\mathrm{\pi }}{3}\right)$ and ${\mathrm{y}}_2 = 3\left(\sin 2\mathrm{\pi t} + \cos 2\mathrm{\pi t}\right)$. The ratio of their amplitudes is

A disc executes S.H.M. about the axis XX' in the plane of the disc as shown in the figure. Its time period of oscillation is:

A spring-block system shown in figure oscillates with a certain time period. If charge q is given to the block and a uniform field E is switched on, then its time period of oscillation is

The period of oscillation of the spring block system shown in the figure is: (assume pulleys and spring to be ideal)

The graph between velocity and acceleration of a particle executing S.H.M. can be

The potential energy of a particle of mass m executing SHM is given by U = A(1 - cos2x), where x is the instantaneous displacement of the particle. The time period of oscillation is

A uniform rod of length l is suspended at $\frac{\mathrm{l}}{4}$ from one end and made to undergo small oscillations. The time period of oscillation is:

A simple pendulum with a metallic bob has a time period T. The bob is now immersed in a nonviscous liquid and the time period is found to be $\sqrt{5}$T. The ratio of the density of the metal to that of liquid is 

A particle executes S.H.M. with time period T. The time period of oscillation of total energy is:

A particle is executing linear simple harmonic motion with an amplitude a and an angular frequency \(\omega\). Its average speed for its motion from extreme to mean position will be

The time period of oscillation of a simple pendulum of length equal to half of the diameter of the earth is about 

The time period of the given spring-mass system is

The equation of simple harmonic motion is given by X = (4 cm)$\sin \left(6.28 t + \frac{5\mathrm{\pi }}{3}\right)$, then maximum velocity of the particle in simple harmonic motion is: 

A spring pendulum is on the rotating table. The initial angular velocity of the table is ${\mathrm{\omega }}_0$ and the time period of the pendulum is ${\mathrm{T}}_0$. Now the angular velocity of the table becomes 2${\mathrm{\omega }}_0$, then the new time period will be:

If vertical spring-mass system dipped in a non-viscous liquid, then:

The displacement x of a particle varies with time t as $\mathrm{x} = \mathrm{A} \sin \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t} + \frac{\mathrm{\pi }}{3}\right)$. Time taken by particle to reach from $\mathrm{x} = \frac{\mathrm{A}}{2}$ to $\mathrm{x} = -\frac{\mathrm{A}}{2}$ will be

Force on a particle F varies with time t as shown in the given graph. The displacement x vs time t graph corresponding to the force-time graph will be
          

The time period of a simple pendulum in a stationary trolley is ${\mathrm{T}}_1$. If the trolley is moving with a constant speed, then time period is ${\mathrm{T}}_2$, then 

A particle executes S.H.M with a frequency of 20 Hz. The frequency with which its potential energy oscillates is:

Two simple pendulums of lengths 1.44 m and 1 m start S.H.M. together in the same phase. They will be in the same phase again after

 A particle is moving along the x-axis. The speed of particle v varies with position x as $\frac{{\mathrm{v}}^2}{144} + \frac{{\mathrm{x}}^2}{9} = 1$. The time period of S.H.M is

A block of mass m is attached to a massless spring having a spring constant k. The other end of the spring is fixed from the wall of a trolley, as shown in the figure. Spring is initially unstretched and the trolley starts moving toward the direction shown. Its velocity-time graph is also shown.

The energy of oscillation, as seen from the trolley is:

A body of mass 20 g is executing S.H.M with amplitude 5 cm. When it passes through equilibrium position its speed is 20 cm/s. Find the distance from equilibrium when its speed becomes 10 cm/s.

For a simple harmonic oscillator, a velocity-time diagram is shown. The angular frequency of oscillation is:

In an S.H.M, when the displacement is one-fourth of the amplitude, the ratio of total energy to the potential energy is