Two particles P and Q start from origin and execute Simple Harmonic Motion along X-axis with same amplitude but with periods 3 seconds and 6 seconds respectively. The ratio of the velocities of P and Q when they meet is -

The angular velocities of three bodies in simple harmonic motion are ${\omega }_1,{\omega }_2,{\omega }_3$ with their respective amplitudes as $A_1,A_2,A_3$. If all the three bodies have same mass and maximum velocity, then

(a) $A_1{\omega }_1=A_2{\omega }_2=A_3{\omega }_3$        (b)  $A_1{{\omega }_1}^2=A_2{{\omega }_2}^2=A_3{A_3}^2$

(b) ${A_1}^2{\omega }_1={A_2}^2{\omega }_2={A_3}^2{\omega }_3$   (d) ${A_1}^2{{\omega }_1}^2={A_2}^2{{\omega }_2}^2=A^2$  

The amplitude of a particle executing SHM is 4 cm. At the mean position the speed of the particle is 16 cm/sec. The distance of the particle from the mean position at which the speed of the particle becomes $8\sqrt{3}\mathrm{cm}/\sec$ will be

The maximum velocity of a simple harmonic motion represented by $y=3 \sin \left(100t+\frac{\mathrm{\pi }}{6}\right)$ is given by

The displacement equation of a particle is $x=3\sin 2t+4\cos 2t$ The amplitude and maximum velocity will be respectively

The instantaneous displacement of a simple pendulum oscillator is given by $x=A \cos \left(\omega t+\frac{\mathrm{\pi }}{4}\right)$ . Its speed will be maximum at time

The amplitude of a particle executing S.H.M. with frequency of 60 Hz is 0.01 m. The maximum value of the acceleration of the particle is

(a)      $144{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$        (b)     $144m/se{c}^2$

c)      $\frac{144}{{\mathrm{\pi }}^2}m/se{c}^2$           (d)     $288{\mathrm{\pi }}^2\mathrm{m}/{\sec }^2$   

A particle moving along the x-axis executes simple harmonic motion, then the force acting on it is given by

What is the maximum acceleration of the particle doing the SHM $y=2\sin \left[\frac{\mathrm{\pi t}}{2}+\varphi \right]$ where 2 is in cm

(a)  $\frac{\mathrm{\pi }}{2}cm/s^2$                 (b)    $\frac{{\mathrm{\pi }}^2}{2}cm/s^2$

(c)    $\frac{\mathrm{\pi }}{4}cm/s^2$               (d)    $\frac{\mathrm{\pi }}{4}cm/s^2$

A particle executes simple harmonic motion along a straight line with an amplitude A. The potential energy is maximum when the displacement is

The potential energy of a particle with displacement X depends as U(X). The motion is simple harmonic, when (K is a positive constant)

A rectangular block of mass m and area of cross-section A floats in a liquid of density ρ. If it is given a small vertical displacement from equilibrium, it undergoes oscillation with a time period T. Then:

The angular velocity and the amplitude of a simple pendulum is $\omega$ and a respectively. At a displacement X from the mean position if its kinetic energy is T and potential energy is V, then the ratio of T to V is 

A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy changes into potential energy, will be

There is a body having mass m and performing S.H.M. with amplitude a. There is a restoring force ,$F=-Kx$ where x is the displacement. The total energy of body depends upon -

The potential energy of a simple harmonic oscillator when the particle is half way to its end point is (where E is the total energy)

A body executes simple harmonic motion. The potential energy (P.E.), the kinetic energy (K.E.) and total energy (T.E.) are measured as a function of displacement x. Which of the following statements is true ?

­­A man measures the period of a simple pendulum inside a stationary lift and finds it to be T sec. If the lift accelerates upwards with an acceleration $\raisebox{1ex}{$\mathrm{g}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$ , then the period of the pendulum will be

The total energy of a particle, executing simple harmonic motion is

The bob of a pendulum of length l is pulled aside from its equilibrium position through an angle $\theta$ and then released. The bob will then pass through its equilibrium position with a speed v, where v equals

A body is executing Simple Harmonic Motion. At a displacement x its potential energy is $E_1$ and at a displacement y its potential energy is $E_2$. The potential energy E at displacement $\left(x+y\right)$ is 

In a simple pendulum, the period of oscillation T is related to length of the pendulum l as

The equation of motion of a particle is $\frac{d^{2}y}{d{t}^2}+Ky=0$ where K is positive constant. The time period of the motion is given by

The kinetic energy of a particle executing S.H.M. is 16 J when it is in its mean position. If the amplitude of oscillations is 25 cm and the mass of the particle is 5.12 kg, the time period of its oscillation is -

A pendulum has time period T. If it is taken on to another planet having acceleration due to gravity half and mass 9 times that of the earth, then its time period on the other planet will be

A simple pendulum hanging from the ceiling of a stationary lift has a time period T₁. When the lift moves downward with constant velocity, then the time period becomes T_2. It can be concluded that 

If the length of a pendulum is made 9 times and mass of the bob is made 4 times, then the value of time period will become

A simple harmonic wave having an amplitude a and time period T is represented by the equation $y=5 \mathrm{sin\pi }\left(t+4\right)$m Then the value of amplitude (a) in (m) and time period  (T) in second are       

The period of a simple pendulum measured inside a stationary lift is found to be T. If the lift starts accelerating upwards with acceleration of g/3 then the time period of the pendulum is

The time period of a simple pendulum of length L as measured in an elevator descending with acceleration $\frac{\mathrm{g}}{3}$ is