Oscillations Te...

 The amplitude of a simple pendulum, oscillating in air with a small spherical bob, decreases from $$10\ cm$$ to $$8\ cm$$ in $$40$$ seconds. Assuming that Stokes law is valid, and ratio of the coefficient of viscosity of air to that of carbon dioxide is $$1.3$$, the time In which amplitude of this pendulum will reduce from $$10\ cm$$ to $$5\ cm$$ in carbon dioxide will be close to (ln $$5=1.601, \ln { 2 }  2=0.693$$)

The potential energy of a particle of mass 10g varies as its displacement from its mean position given by $$U = 3x^2 + 3$$, then, the particle performs a

A particle executes SHM along x axis and is at the mean position at t=0. What is its velocity at its mean position. The amplitude of SHM is 5 cm and angular frequency is 2 rad/s:

A particle executes SHM given by the equation $$x = 4 sin (2 \pi t +\pi/4)$$, what will be the velocity of the particle at t = (1/8)th sec; 

The potential energy of a particle is directly proportional to its linear displacement from its mean position. Then, the particle performs a

At what position along a straight line will the velocity be zero for a particle executing SHM

A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that is at a distance $$\dfrac{2A}{3}$$ from equilibrium position. The new amplitude of the motion is:

What type of curve do we get, if $$x^2$$ and $$v^2$$ are plotted for a particle executing SHM, x and v are the position and velocity of the particle:

A particle executes SHM from its mean position at t=0 with an amplitude A, what will be its velocity at x=A/2, in its forward motion towards the extreme:

The time period of oscillation of magnet in a vibrating magnetometer is $$1.5$$ sec. The time period of oscillation of another magnet similar in size and mass but having one-fourth the magnetic moment than that of the first magnet oscillating at the same place will be