Simple Harmonic Motion · Physics · COMEDK
MCQ (Single Correct Answer)
A body is executing SHM. When its displacements from the mean position are $$4 \mathrm{~cm}$$ and $$5 \mathrm{~cm}$$ it has velocity $$10 \mathrm{~cms}^{-1}$$ and $$8 \mathrm{~cms}^{-1}$$ respectively. Its periodic time $$\mathrm{t}$$ is
A circular disc of mass $$20 \mathrm{~kg}$$, having radius $$10 \mathrm{~cm}$$ is suspended by a wire attached to its centre. The wire is twisted by rotating the disc and released. The time period of torsional oscillations is found to be $$1 \mathrm{~s}$$. The torsional spring constant of the wire is
A particle executes a simple harmonic motion of amplitude $$\mathrm{A}$$. The distance from the mean position at which its kinetic energy is equal to its potential energy is
The phase difference between displacement and velocity of a particle in simple harmonic motion is
A particle is performing simple harmonic motion. Equation of its motion is $$x = 5\sin \left( {4t - {\pi \over 6}} \right),x$$ being the displacement from mean position. Velocity (in ms$$^{-1}$$) of the particle at the instant when its displacement is 3, will be
The function $$y = \log \omega t$$ can represent
Two spring of force constant $$k_1$$ and $$k_2$$ are configured as the figure given below
The angular frequency of this configuration is
A particle executes a linear SHM with an amplitude of 4 cm. At the mean position the velocity of the particle is 10 cm/s. What is the displacement of the particle when its speed becomes 5 cm/s?