JEE Main || Mechanics || Super Quiz 12

A simple harmonic oscillator of angular frequency 2 rad s^-1 is acted upon by an external force F= sint N. If the oscillator is at rest in its equilibrium position at t=0, its position at later times is proportional to:

A block of mass 0.1 kg is connected to an elastic spring of spring constant 640 Nm^-1 and oscillates in a damping medium of damping constant 10^-2 kg s^-1. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to:

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 carbondioxide will be close to

Two springs, of force constants k₁ and k₂, are connected to a mass m as shown. The frequency of oscillation of the mass is f. If both k₁ and k₂ are made four times their original values, the frequency of oscillation becomes

A particle at the end of a spring executes Simple Harmonic Motion with a period t₁, while the corresponding period for another spring is t₂. If the period of oscillation with the two springs in series is t, then

Starting from the origin a body oscillates simple harmonically with a period of 2 s. After what time will its kinetic energy be 75% of the total energy ?

Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x–axis. Their mean position is separated by distance X₀ (X₀ > A). If the maximum separation between them is (X₀ + A), the phase difference between their motion is

A particle executive a simple harmonic motion has a period of 6 sec. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance equal to half its amplitude, the distance being measured from its equilibrium position is

A particle is executing simple harmonic motion with a time period T. At time t=0, it is at its position of equilibrium. The kinetic energy - time graph of the particle will look like: