Simple Harmonic Motion Test - 14

Suppose that a pendulum clock is carried to a depth of 32 km inside the earth (Radius R = 6400 km). To have the correct time from the clock, by what percentage the effective length of the pendulum should be changed?

A particle is executing SHM with amplitude A and angular frequency ω. The time taken by the particle to move from x = 0 to x = A/2 is

For a particle executing SHM, the graph between its momentum and displacement will be

A particle executes SHM of amplitude 10 m and period 4 s along a straight line. The velocity of the particle at a distance of 6 m from the mean position is: 

A pendulum oscillates about its mean position C. The position where the speed of the bob becomes maximum is

The position x (in centimeter) of a simple harmonic oscillator varies with time t (in second) as x = 2cos (0.5πt + π/3) The magnitude of the maximum acceleration of the particle in cm/s² is:

A horizontal platform is executing simple harmonic motion in the vertical direction with frequency f. A block of mass m is placed on the platform. What is the maximum amplitude of the SHM, so that the block is not detached from it?

The equation of a particle executing simple harmonic motion is y = 0.4 sin (2πt + π/3) (where t is in seconds and y is in meters). The initial phase of the particle is:

The equation of a particle executing simple harmonic motion is y = 2√2 sin 314t. Displacement y from the mean position where acceleration becomes zero is: (y is in cm and t is in second) 

A body is placed on a horizontal platform that is undergoing vertical SHM. If the amplitude of oscillation is 40 cm, then the least period of oscillation for which an object placed over the platform is not detached from it is: