AIIMS Physics Oscillations Class 11 Questions

The formula to calculate the equivalent spring constant is given by,

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

Find the maximum tension in the spring if initially spring at its natural length when block is released from rest.

In the case of damped oscillation, the amplitude keeps on decreasing that implies the graph for V versus x is,

The time after which the energy will become half of initial maximum value in damped forced oscillation is calculated as, 1/√2 = e^(-bt/m) ln√2 = bt/m t = m/b × 1/2 ln 2

The expression of force is given as, F = mω²A = μmg So, A = μg/ω² And ω = √(K/m) So, ω = √(100/1.5) Substitute the values.

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

In damped oscillation, the amplitude will decrease so the graph of V v/s x will be as,

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

The formula for the time period of simple pendulum of length is, T = 2π√(l/g) T ∝ √l Now, √(ΔT/T) = 1/2 Δl/l √(ΔT/T) = 1/2 × 3% = 1.5%

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).

It is given that, m = 4 kg, k = 800 N/m, E = 4 J. For simple harmonic motion, E = ½kA². Calculate A.

Two identical pendulums are suspended from a common point. The masses of the pendulums are m and q, and the lengths are L. Calculate the time period of small oscillations.

In forced vibration m = 10 gm, f = 100 Hz and driver force F = 100 cos (20πt), 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

Kinetic energy and potential energy completes two vibration in a time during which SHM completes one vibration. Thus frequency of potential energy and kinetic energy is double than that of SHM.

The maximum velocity of the particle is calculated as, vₘₐₓ = a(2πf) = 0.1 cm(2π × 300 Hz) = 60π cm/s

Amplitude of damped oscillator at an instant t is, A = A₀e^{-bt/2m}. When t = 2 s, A = A₀/3. When t = 6 s, A = A₀/n. Find n.

v₁ = \( \frac{dy₁}{dt} \) = 0.1 × 100π cos(100πt + \( \frac{π}{3} \) ) v₂ = \( \frac{dy₂}{dt} \) = -0.1π sin πt = -0.1π cos(πt + \( \frac{π}{2} \) ) Phase difference = \( \frac{π}{3} \) - \( \frac{π}{2} \) = -\( \frac{π}{6} \)

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)

The force of the particle with respect to an observer is given as, F = -k[x + (v₀ - v₀)t] F = -kx Thus, it represents the simple harmonic motion.

If amplitude of simple pendulum increases then the motion of pendulum is oscillatory but not simple harmonic because for larger amplitude θ is large.

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