A body describes simple harmonic motion with an amplitude of 5 cm and a period of 0.2 s. What is the velocity of the body when the displacement is 3 cm?

A mass attached to a spring is free to oscillate, with angular velocity ω, in a horizontal plane without friction or damping. It is pulled to a distance $x_0$ and pushed towards the centre with a velocity $v_0$ at time t = 0. The amplitude of the resulting oscillations is:

$$\begin{gathered} \left(1\right) \sqrt{\left(2{{\mathrm{x}}_0}^2+\frac{{{\mathrm{v}}_0}^2}{{\mathrm{\omega }}^2}\right)} \\ \left(2\right) \sqrt{\left({{\mathrm{x}}_0}^2+\frac{{{\mathrm{v}}_0}^2}{{\mathrm{\omega }}^2}\right)} \\ \left(3\right) \sqrt{\left({{\mathrm{x}}_0}^2+\frac{{{\mathrm{v}}_0}^2}{2{\mathrm{\omega }}^2}\right)} \\ \left(4\right) \sqrt{\left({{\mathrm{x}}_0}^2+\frac{{{\mathrm{v}}_0}^2}{{\mathrm{\pi \omega }}^2}\right)} \end{gathered}$$

Identify the correct definition:

The displacement of a particle executing S.H.M. is given by x = 0.01sin100$\pi$(t + 0.05). The time period is:

A boy is swinging in a swing. If he stands, the time period will:

The time period of a simple pendulum in a freely falling lift will be:

If the effective length of a simple pendulum is equal to the radius of the earth (R), the time period will be:

A body executing S.H.M. along a straight line has a velocity of 3 ms^-1 when it is at a distance of 4 m from its mean position and 4 ms^-1 when it is at a distance of 3 m from its mean position. Its angular frequency and amplitude are:

The frequency of oscillation of a mass m suspended by a spring is ${\mathrm{\nu }}_1$. If the length of the spring is cut to one third, then the same mass oscillates with a frequency ${\mathrm{\nu }}_2$_, then:

The equation of simple harmonic motion may not be expressed as (each term has its usual meaning):

If a particle is executing simple harmonic motion, then the acceleration of the particle:

What is the phase difference between the acceleration and the velocity of a particle executing simple harmonic motion?

The shape of the graph plotted between the velocity and the position of a particle executing simple harmonic motion is:

If a particle is executing simple harmonic motion with time period T, then the time period of its total mechanical energy is:

Select the wrong statement about simple harmonic motion.

A particle is executing SHM with time period T starting from the mean position. The time taken by it to complete $\frac{5}{8}$ oscillations is:

A particle is executing S.H.M. between x = $\pm$ A. The time taken to go from 0 to $\frac{A}{2}$ is T₁ and to go from $\frac{A}{2}$ to A is T_2, then:

For a particle executing simple harmonic motion, the amplitude is A and the time period is T. The maximum speed will be:

A particle is executing S.H.M with an amplitude A and has maximum velocity v_0, its speed at displacament $\frac{3\mathrm{A}}{4}$ will be:

Two particles executing SHM of the same frequency meet at x = +a/2 while moving in opposite directions. The phase difference between the particles is:

The displacements of two particles executing SHM on the same line are given as
${\mathrm{y}}_1=asin\left(\frac{\mathrm{\pi }}{2}\mathrm{t}+\mathrm{\varphi }\right)$ and ${\text{y}}_2=\mathrm{bsin}\left(\frac{2\mathrm{\pi }}{3}\mathrm{t}+\mathrm{\varphi }\right)$. The phase difference between them at t=1 sec is:

For a particle showing motion under the force F= -5 (x - 2)², the motion is:

For a particle showing motion under the force F= -5(x - 2), the motion is:

Two identical springs have the same force constant 73.5 Nm^-1. The elongation produced in each spring in three cases shown in Figure-1, Figure- 2 and Figure- 3 are: (g = 9.8 ms^-2)

A particle of mass 10 g is undergoing SHM of amplitude 10 cm and period 0.1 sec. The maximum value of the force on the particle is about:

Two masses m₁=1 kg and m₂=0.5 kg are executing SHM together suspended by a massless spring of spring constant 12.5 Nm^-1. When masses are in equilibrium, m₁ is removed without disturbing the system. The new amplitude of oscillation will be: