AIIMS Physics Oscillations Class 11 Questions

Time period of oscillation for given combination will be:

An oscillator circuit consists of an inductance of 0.5 mH and a capacitor of 20 μF. The resonant frequency of the circuit is nearly

In damped oscillation graph between velocity and position will be

$$mx^2 - bx + k = 0$$ Find time after which to the energy will become half of initial maximum value in damped force oscillation.

Maximum amplitude of SHM so block A will not slip on block B, $\text{K} = 100\text{ N/m}$

A body oscillates with a simple harmonic motion having amplitude 0.05 m. At a certain instant, its displacement is 0.01 m and acceleration is 1.0 m/s². The period of oscillation is

The graph between velocity and position for a damped oscillation will be :-

Assertion: Positive feedback is essential for converting a transistor into an oscillator. Reason: Positive feedback works between cut-off and saturation region.

Assertion : The percentage change in time period is 1.5%, if the length of simple pendulum increases by 3%. Reason : Time period is directly proportional to length of pendulum.

Assertion: In simple harmonic motion, the motion is to and fro and periodic. Reason: Velocity of the particle in simple harmonic motion $(v) = \omega\sqrt{k^2 - x^2}$ (where x is the displacement and k is spring factor).

A mass of $4\text{ kg}$ suspended from a spring of force constant $800\text{ N m}^{-1}$ executes simple harmonic oscillations. If the total energy of the oscillator is $4\text{ J}$, the maximum acceleration (in $\text{m s}^{-2}$) of the mass is

Two small balls, each having equal positive charge $Q$ are suspended by two insulating strings of equal length $L$ from a hook fixed to a stand. If mass of each ball = $m$ & total angle between the two strings is $60^\circ$, then find the charge on each ball.

In forced vibration $m = 10\text{ gm}$, $f = 100\text{ Hz}$ and driver force $F = 100\cos(20\pi t)$, then find the amplitude of particle

A particle performs SHM on x-axis with amplitude A and time period T. The time taken by the particle to travel a distance A/5 starting from rest is

A particle of mass is executing oscillations about the origin on the x-axis. Its potential energy is V(x)=k|x|^3, where k is a positive constant. If the amplitude of oscillation is a, then its time period T is

The velocity vector v and displacement vector x of a particle executing SHM are related as \( \frac{dv}{dx} = -\omega^2 x \) with the initial condition \( V = v₀ \) at \( x = 0 \). The velocity v, when displacement is x, is

What do you conclude from the graph about the frequency of KE, PE and SHM.

When a sound wave of frequency $300\text{ Hz}$ passes through a medium, the maximum displacement of a particle of the medium is $0.1\text{ cm}$. The maximum velocity of the particle is equal to

The amplitude of a damped oscillator becomes 1/3rd in 2 seconds. If its amplitude after 6 seconds is $\frac{1}{n}$ times the original amplitude, the value of $n$ is:

Two simple harmonic motions are represented by the equations $y_1 = 0.1\sin\left(100\pi t + \frac{\pi}{3}\right)$ and $y_2 = 0.1\cos \pi t$. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is:

A simple pendulum has time period T₁. The point of suspension is now moved upward according to the relation y = Kt², (K = 1 m/s²) where y is the vertical displacement. The time period now becomes T₂. The ratio of T₁²/T₂² is (g = 10 m/s²)

Assertion : The force acting on a particle moving along x-axis is $F = -k(x + v_0t)$, where $k$ is a constant. Reason : To an observer moving along x-axis with constant velocity $v_0$, it represents S.H.M.

Assertion : If amplitude of simple pendulum increases then the motion of pendulum is oscillatory but not simple harmonic. Reason: For larger amplitude θ is large and then sin θ ≠ θ, so the motion is no longer SHM.

Assertion: If the amplitude of a simple harmonic oscillator is doubled, its total energy becomes four times. Reason: The total energy is directly proportional to the square of the amplitude of vibration of the harmonic oscillator.

Assertion: A particle executing simple harmonic motion comes to rest at the extreme positions. Reason: The resultant force on the particle is zero at these positions.

A long mass m is fall from the height h on the scale pan hung from the spring as shown. If the spring constant is k and the mass of pan is zero and the mass m does not bounce relative to the pan, the amplitude of the vibration is

The angular amplitude of a simple pendulum is θ₀. The maximum tension in its string will be

The displacement of a particle executing SHM is given by y = 0.25 sin 200t cm. The maximum speed of the particle is

A spring-mass system performs oscillations with a time period T. If the mass is increased by m, then the time period becomes \( \frac{5}{4} T \). The ratio of M/m is

The frequency of oscillations of a mass m connected horizontally by a spring of spring constant k is 4 Hz. When the spring is replaced by two identical springs as shown in figure. Then the effective frequency is,

Assertion : The graph of potential energy and kinetic energy of a particle in SHM with respect to position is a parabola. Reason : Potential energy and kinetic energy do not vary linearly with position.

The velocity vector \( v \) and displacement vector \( x \) of a particle executing SHM are related as \( \frac{dv}{dx} = -\omega^2 x \) with the initial condition \( v = v_0 \) at \( x = 0 \). The velocity \( v \), when displacement is \( x \), is

In an oscillating system, a restoring force is a must. In an L⁻C circuit, restoring force is provide by

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude 'a' and time period 'T'. The speed of the pendulum at x = a/2 will be

Assertion: A undamped spring-mass system is simplest free vibration system. Reason: It has three degrees of freedom.

A 4 kg roller is attached to a massless spring of spring constant k = 100 N/m. It rolls without slipping along a frictionless horizontal road. The roller is displaced from its equilibrium position by 10 cm and then released. Its maximum speed will be

A particle moving about its equilibrium position with equation y = −ax − bt. Interpret the condition it will always perform the SHM.

If maximum speed of a particle in SHM is given by Vₘₐₓ, what is its average speed?

Which of the following equation does not represent a SHM?

In simple harmonic motion, loss of kinetic energy is proportional to

Let T₁ and T₂ be the time periods of springs A and B when mass M is suspended from one end of each spring. If both springs are taken in series and the same mass M is suspended from the series combination, the time period is T, then

A particle executes simple harmonic motion of period T and amplitude A along a rod AB of length 2l. The rod AB itself executes simple harmonic motion of the same period and amplitude in a direction perpendicular to its length. Initially, both the particle and the rod are in their mean positions. The path traced out by the particle will be

A ball is suspended by a thread of length L at the point O on a wall which is inclined to the vertical by α. The thread with the ball is displaced by a small angle β away from the vertical and also away from the wall. If the ball is released, the period of oscillation of the pendulum when β > α will be

Assertion : The bob of a simple pendulum is a ball full of water, if a fine hole is made in the bottom of the ball, the time period first increases and then decreases. Reason : As water flows out of the bob the weight of bob decreases.

Assertion: In a SHM, kinetic and potential energies become equal when the displacement is 1/√2 times the amplitude. Reason: In SHM, kinetic energy is zero when potential energy is maximum.

A large horizontal surface moves up and down in S.H.M. with an amplitude of 1 cm. If a mass of 10 kg (which is placed on the surface) is to remain continuously in contact with it, the maximum frequency of S.H.M. will be

Which of the following functions represents a simple harmonic oscillation?

Two springs are connected to a block of mass M placed on a frictionless surface as shown below. If both the springs have a spring constant k, the frequency of oscillation of the block is

Two springs of force constants k and 2k are connected to a mass as shown in figure. The frequency of oscillation of the mass is

Assertion : The amplitude of an oscillating pendulum decreases gradually with time. Reason : The frequency of the pendulum decreases with time.