JEE Main Simple Harmonic Motion Practice Questions With Solutions

A particle executes S.H.M. of amplitude A along x-axis. At t = 0, the position of the particle is $x=\frac{A}{2}$x = \frac{A}{2} and it moves along positive x-axis. The displacement of particle in time t is $x=A\sin (wt+\delta )$x = A sin ⁡ \left(\right. w t + \delta \left.\right), then the value of $\delta$\delta will be

For particle P revolving round the centre O with radius of circular path $\mathrm{r}$r and angular velocity $\omega$\omega, as shown in below figure, the projection of OP on the $x$x-axis at time $t$t is

A mass $m$m is attached to two strings as shown in figure. The spring constants of two springs are ${\mathrm{K}}_1$K_{1} and ${\mathrm{K}}_2$K_{2}. For the frictionless surface, the time period of oscillation of mass $m$m is :

Choose the correct length (L) versus square of the time period (${\mathrm{T}}^2$T^{2}) graph for a simple pendulum executing simple harmonic motion.

The maximum potential energy of a block executing simple harmonic motion is $25\mathrm{J}$25 \textrm{ } J. A is amplitude of oscillation. At $\mathrm{A}/2$A / 2, the kinetic energy of the block is

For a simple harmonic motion in a mass spring system shown, the surface is frictionless. When the mass of the block is $1\mathrm{kg}$1 \textrm{ } kg, the angular frequency is ${\omega }_1$\omega_{1}. When the mass block is $2\mathrm{kg}$2 \textrm{ } kg the angular frequency is ${\omega }_2$\omega_{2}. The ratio ${\omega }_2/{\omega }_1$\omega_{2} / \omega_{1} is

A particle executes simple harmonic motion between $x=-A$x = - A and $x=+A$x = + A. If time taken by particle to go from $x=0$x = 0 to $\frac{A}{2}$\frac{A}{2} is 2 s; then time taken by particle in going from $x=\frac{A}{2}$x = \frac{A}{2} to A is

T is the time period of simple pendulum on the earth's surface. Its time period becomes $x$x T when taken to a height R (equal to earth's radius) above the earth's surface. Then, the value of $x$x will be :

The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination $\alpha$\alpha, is given by :

Assume there are two identical simple pendulum clocks. Clock - 1 is placed on the earth and Clock - 2 is placed on a space station located at a height h above the earth surface. Clock - 1 and Clock - 2 operate at time periods 4 s and 6 s respectively. Then the value of h is -

(consider radius of earth $R_E=6400\mathrm{km}$R_{E} = 6400 \textrm{ } km and $\mathrm{g}$g on earth $10\mathrm{m}/{\mathrm{s}}^2$10 \textrm{ } m / s^{2} )