Question 1

A block of mass M is attached to a horizontal spring and oscillates in simple harmonic motion with amplitude A1. When the block crosses the equilibrium position, a smaller mass m is gently placed on it, and thereafter the two move together. The new amplitude is A2. What is the value of the

Accepted Answer

((M + m)/M)^(1/2). The correct option is derived from the conservation of energy principle in simple harmonic motion. When the smaller mass m is added to the block of mass M at the equilibrium position, the total mass increases, which reduces the amplitude of oscillation. The new amplitude A2 is related to the original amplitude A1 by the ratio of the square root of the total mass to the original mass, leading to the expression A1/A2 = (M + m)/M.

Question 2

An ideal gas is trapped in a vertical cylinder of cross-sectional area A by a frictionless piston of mass M. When the gas volume is V0 and its pressure is P0, the piston is in equilibrium. If the piston is given a small displacement from this position and then let go, and the entire setup

Accepted Answer

(1/2π)√(A²γP0/(MV0)). For an adiabatic small displacement x, dP = -(gamma*P0/V0)*A*x, so restoring force = -(gamma*P0*A^2/V0)*x. Then omega^2 = gamma*P0*A^2/(M*V0) and f = (1/2pi)*sqrt(A^2*gamma*P0/(M*V0)).

Question 3

A mass m is connected to a spring with spring constant k, giving it a natural angular frequency ω0. If an external driving force varying as cos(ωt) acts on the system, the oscillator’s displacement will be proportional to

Accepted Answer

1/(m(ω0² - ω²)). For a forced oscillator (neglecting damping), steady-state displacement amplitude is proportional to 1/(m*(w0^2 - w^2)).

Question 4

A particle undergoing simple harmonic motion has kinetic energy K0. The maximum potential energy and the total energy are, respectively,

Accepted Answer

K0 and K0. Taking the given K0 as the (maximum) kinetic energy, the total energy equals K0, and the maximum potential energy also equals the total energy, so both are K0.

Question 5

A simple pendulum is suspended from the roof of a lift that is initially at rest, and its period is T. The lift then moves upward such that the distance it covers is given by y = t², where y is in metres and t is in seconds. Taking g = 10 m/s², what will be the new time period of the pendu

Accepted Answer

√(5/6) T. From y = t^2, acceleration = d^2y/dt^2 = 2 m/s^2 upward, so effective gravity = g + a = 12 m/s^2. New period T' = T*sqrt(g/g_eff) = T*sqrt(10/12) = sqrt(5/6) T.

Question 6

A damped oscillator’s amplitude falls to 0.9 of its initial value in 5 s. After an additional 10 s, what fraction of the initial amplitude will it have, if the factor is denoted by α?

Accepted Answer

0.729. Amplitude decays as A = A0*e^(-bt). In each 5 s interval it multiplies by 0.9. After 15 s total (3 intervals): (0.9)^3 = 0.729 of the initial amplitude.

Question 7

A block of mass 1 kg is connected to a spring of force constant 600 N/m and is at rest on a frictionless horizontal table, with the other end of the spring fixed to a wall. Another block of mass 0.5 kg moves along the table toward the first block with speed 3 m/s. If the two blocks undergo

Accepted Answer

5 cm, π/10 s. After collision v = (0.5*3)/1.5 = 1 m/s, combined mass 1.5 kg. omega = sqrt(600/1.5) = 20 rad/s. A = v/omega = 1/20 = 0.05 m = 5 cm; T = 2pi/20 = pi/10 s.

Question 8

A damped pendulum has a time period of 1 s and is continuously losing energy. At one instant, its energy is 45 J. After it has executed 15 complete oscillations, its energy falls to 15 J. The damping constant (in s⁻¹) is:

Accepted Answer

1/15 ln3. The damping constant can be determined using the relationship between the initial and final energy of the damped pendulum over a certain number of oscillations. The formula for energy decay in a damped system shows that the energy decreases exponentially, and by applying this to the given energies and oscillations, we find that the damping constant is correctly calculated as 1/15 ln3.

Question 9

A particle of mass 0.1 kg performs simple harmonic motion with amplitude 0.1 m. As it crosses the equilibrium position, its kinetic energy is 8 × 10⁻³ J. Find the equation of motion of the particle, given that the initial phase is 45°.

Accepted Answer

y = 0.1 sin (4t + π/4). (1/2)*0.1*w^2*(0.1)^2 = 8*10^-3 -> w^2 = 16 -> w = 4 rad/s. Amplitude is 0.1 m, so y = 0.1 sin(4t + pi/4).

Question 10

A simple pendulum is suspended by a uniform wire of cross-sectional area A and has time period T. If an extra mass M is attached to its bob, the new time period becomes TM. For the wire material with Young’s modulus Y, the value of 1/Y is:

Accepted Answer

[(TM/T)² - 1] A / Mg. The correct option relates the change in time period due to the added mass to the material properties of the wire, specifically Young's modulus. The formula shows how the increase in time period affects the elongation of the wire, which is directly linked to its Young's modulus, thus providing a relationship between the mechanical properties and the pendulum's behavior.

Question 11

A coin rests on a horizontal plate that executes vertical simple harmonic motion with angular frequency ω. If the oscillation amplitude is slowly increased, the coin will first lose contact with the plate

Accepted Answer

when the amplitude reaches g/ω². The coin will lose contact with the plate when the upward acceleration of the plate equals the acceleration due to gravity. This occurs when the amplitude of the oscillation reaches g/ω², as this is the point where the maximum upward acceleration of the plate is insufficient to keep the coin pressed against it.

Question 12

A simple pendulum bob oscillates with period t while immersed in water. If the same bob oscillates in air with period t0, and the viscous resistance of water is neglected, then for a bob whose density is (4/3) × 1000 kg/m³, the relation between t and t0 is

Accepted Answer

t = 2t0. Effective gravity in water g' = g(1 - rho_water/rho_bob) = g(1 - 1000/1333.3) = g/4. Since T is proportional to 1/sqrt(g), t/t0 = sqrt(g/g') = sqrt(4) = 2, so t = 2*t0.

Question 13

A particle begins SHM from the mean position (origin) with a time period of 2 s. After how much time will its kinetic energy become 75% of the total mechanical energy?

Accepted Answer

1/6 s. From mean position x = A sin(wt). KE = 0.75 E means PE = 0.25 E -> x = A/2 -> sin(wt)=1/2 -> wt = pi/6. With T = 2 s (w = pi), t = (pi/6)/pi = 1/6 s.

Question 14

For a particle executing simple harmonic motion, what is the ratio of the kinetic energy at the equilibrium position to the potential energy when its displacement is equal to half the amplitude?

Accepted Answer

4/1. KE at equilibrium = total energy = (1/2)kA^2. PE at x=A/2 = (1/2)k(A/2)^2 = (1/8)kA^2. Ratio = (1/2)/(1/8) = 4/1.

Question 15

A particle performs simple harmonic motion under a force F1, and its period is 4/5 s. When the force is replaced by F2, the period becomes 3/5 s. If F1 and F2 act together on the particle in the same direction, what is the resulting period of oscillation in seconds?

Accepted Answer

12/25. k ~ 1/T^2, and combining forces adds the constants: 1/T^2 = 1/T1^2 + 1/T2^2 = (5/4)^2 + (5/3)^2 = 25/16 + 25/9. This gives T^2 = 144/625, so T = 12/25 s.

Question 16

A mass attached to a spring performs vertical oscillations. The block experiences a damping force from the surrounding fluid given by Fd = -bv, where b is a positive constant. The value of b depends on:

Accepted Answer

all of the above. The damping force coefficient b is influenced by the viscosity of the medium, which affects resistance, as well as the dimensions and geometry of the block, which determine how the block interacts with the fluid. Therefore, all these factors collectively influence the value of b.

Question 17

A simple pendulum of length l is pulled to a maximum angular displacement θ. The greatest kinetic energy of its bob of mass m is

Accepted Answer

mg l (1 - cos θ). The greatest kinetic energy of the pendulum bob occurs at the lowest point of its swing, where all the potential energy converted from the height is transformed into kinetic energy. The potential energy at the maximum height is given by mg l (1 - cos θ), which equals the kinetic energy at the lowest point.

Question 18

A tiny charged sphere of charge q is suspended by a string of length l between two parallel plates. Its oscillation period without any electric field is T0. When the plates are given charge, the period becomes T. What is the value of the ratio T/T0?

Accepted Answer

None of these. The correct option is 'None of these' because the oscillation period of the charged sphere in an electric field is influenced by both gravitational and electric forces, leading to a more complex relationship than those presented in the other options.

Question 19

At t = 0, a capacitor carrying its full initial charge q0 is joined to an inductor of self-inductance L. The instant when the energy is shared equally by the electric field and the magnetic field is

Accepted Answer

(π/4)√(LC). Charge q = q0*cos(wt) with w = 1/sqrt(LC); capacitor (electric) energy is proportional to q^2. Equal sharing needs q^2 = q0^2/2, i.e. cos^2(wt) = 1/2, so wt = pi/4 and t = (pi/4)*sqrt(LC).

Question 20

In an LC oscillator, the capacitor attains a peak charge of Q. What is the capacitor charge when the energy is shared equally by the electric field of the capacitor and the magnetic field of the inductor?

Accepted Answer

Q/√2. Total energy = Q^2/(2C). Equal sharing means capacitor energy = Q^2/(4C) = q^2/(2C), giving q^2 = Q^2/2, so q = Q/sqrt(2).

Question 21

A particle of mass 0.1 kg performs simple harmonic motion with amplitude 0.1 m. At the mean position, its kinetic energy is 18 × 10⁻³ J. If the initial phase of the oscillation is 45°, the particle’s equation of motion is given by:

Accepted Answer

x = 0.1 sin(6t + π/4). KE at mean = (1/2)m*omega^2*A^2 = 18e-3 gives (1/2)(0.1)(0.01)omega^2 = 0.018, so omega^2 = 36, omega = 6 rad/s. With initial phase 45 deg, x = 0.1 sin(6t + pi/4).

Question 22

A spring having spring constant k is divided into two segments A and B such that their lengths are in the ratio l_A: l_B = 2: 3. The spring constant of segment A is

Accepted Answer

5k/2. The spring constant of a segment is inversely proportional to its length. Since segment A is shorter than segment B, it will have a higher spring constant. Given the ratio of lengths, the spring constant for segment A can be calculated as 5k/2.

Question 23

A pendulum clock is 12 s slow per day at 40°C, but it is 4 s fast per day at 20°C. The temperature at which it keeps accurate time, and the linear expansion coefficient (α) of the pendulum rod material, are respectively:

Accepted Answer

25°C; α = 1.85 × 10^−5/°C. The pendulum clock's timekeeping is affected by temperature changes, with the clock being slow at higher temperatures and fast at lower temperatures. The correct option indicates that at 25°C, the effects of thermal expansion balance out, allowing the clock to keep accurate time, and the linear expansion coefficient reflects the material's response to temperature changes.

Question 24

A particle at the end of a spring executes S.H.M. 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

Accepted Answer

T² = t1² + t2². For springs in series 1/k = 1/k1 + 1/k2. Since T^2 is proportional to 1/k, T^2 = t1^2 + t2^2.

Question 25

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

Accepted Answer

independent of x. In simple harmonic motion, the total energy is the sum of kinetic and potential energy, both of which vary with position. However, when combined, these energies result in a constant total energy that does not depend on the displacement from the equilibrium position.

Question 26

A particle of mass m is attached to a spring (of spring constant k) and has a natural angular frequency ω0. An external force F(t) proportional to cos ωt (ω ≠ ω0) is applied to the oscillator. The displacement of the oscillator will be proportional to

Accepted Answer

1 / [m(ω² - ω0²)]. The steady-state amplitude of a driven oscillator is F0/[m(omega0^2 - omega^2)], so the displacement is proportional to 1/[m(omega^2 - omega0^2)] (the sign is just a phase). The stored choice puts m in the numerator, which is wrong; mass must appear in the denominator.

Question 27

In forced oscillation of a particle the amplitude is maximum for a frequency ω1 of the force while the energy is maximum for a frequency ω2 of the force; then

Accepted Answer

ω1 < ω2. In forced oscillation, the amplitude is maximized at the resonance frequency, which is lower than the frequency at which energy is maximized due to damping effects. When damping is small, the system can respond more effectively at a lower frequency, leading to the condition where ω1 is less than ω2.

Question 28

Two simple harmonic motions are represented by the equations y1 = 0.1 sin(100πt + π/3)
y2 = 0.1 cos πt. The phase difference of the velocity of particle 1 with respect to the velocity of particle 2 is

Accepted Answer

-π/6. Velocity phase of 1 = pi/3 + pi/2; velocity phase of 2 (from y2=cos = sin(+pi/2)) = pi/2 + pi/2. Difference = (5pi/6) - (pi) = -pi/6.

Question 29

The function sin²(ωt) represents

Accepted Answer

a periodic, but not simple harmonic motion with a period π/ω. Writing sin^2(wt) = (1 - cos 2wt)/2 shows it is periodic with period pi/w, but it is not a simple harmonic motion (its mean position is offset and it is not of the pure form A sin/cos about the origin). Hence it is periodic, not SHM, with period pi/w.

Question 30

The bob of a simple pendulum is a spherical hollow ball filled with water. A plugged hole near the bottom of the oscillating bob gets suddenly unplugged. During observation, till water is coming out, the time period of oscillation would

Accepted Answer

first increase and then decrease to the original value. As water drains, the centre of mass of the water (and hence the effective pendulum length) first lowers then rises back to the centre, so the period first increases and then decreases to its original value.

Question 31

If a simple harmonic motion is represented by d²x/dt² + αx = 0, its time period is

Accepted Answer

2π/√α. Comparing d^2x/dt^2 + alpha x = 0 with d^2x/dt^2 + omega^2 x = 0 gives omega = sqrt(alpha). Time period T = 2pi/omega = 2pi/sqrt(alpha).

Question 32

The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is 4.4 m/s. The period of oscillation is

Accepted Answer

0.01 s. The maximum velocity in simple harmonic motion is given by the formula v_max = Aω, where A is the amplitude and ω is the angular frequency. Given the amplitude and maximum velocity, we can find the angular frequency and then use it to calculate the period, which is the inverse of the frequency.

Question 33

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?

Accepted Answer

1/6 s. KE is 75% of total when PE is 25%, i.e. x = A/2. Starting from the mean position x = A sin(wt), so sin(wt)=1/2, wt = pi/6. With w = 2*pi/T = pi, t = (pi/6)/pi = 1/6 s.

Question 34

A particle of mass point moves along the x-axis with displacement given by x = x0 cos(ωt − π/4). If its acceleration is expressed in the form a = A cos(ωt + δ), then

Accepted Answer

A = x0ω², δ = 3π/4. The acceleration of the particle can be derived from the displacement function by taking the second derivative with respect to time. This results in an expression that shows the amplitude as A = x0ω² and a phase shift of δ = 3π/4, indicating that the acceleration is out of phase with the displacement.

Question 35

Two particles move along the x-axis in simple harmonic motion with the same amplitude A and angular frequency ω. Their equilibrium positions are separated by a distance X0, where X0 > A. If the greatest distance between the particles is X0 + A, what is the phase difference between their os

Accepted Answer

π/3. The phase difference of π/3 indicates that when one particle is at its maximum displacement, the other is at a position that allows them to be separated by the maximum distance of X0 + A. This relationship arises from the properties of simple harmonic motion and the geometry of their oscillations.

Question 36

A block of mass M is attached to a horizontal spring and performs simple harmonic motion with amplitude A1. When it crosses the equilibrium position, a small mass m is gently placed on it so that the two move together thereafter with amplitude A2. What is the ratio A1/A2?

Accepted Answer

((M + m)/M)^(1/2). The correct option reflects the conservation of energy and the relationship between the amplitudes before and after the mass m is added. When the small mass m is placed on the block, the total mass increases, which reduces the amplitude of the system due to the increased inertia, leading to the derived ratio of amplitudes as ((M + m)/M)^(1/2).

Question 37

A pendulum is said to have an average life τ if its amplitude remains appreciable only up to the time when it falls to 1/e of its initial value, i.e. from t=0 to t=τ s. For a spherical bob undergoing small damping due to a viscous resistive force proportional to its speed, with proportiona

Accepted Answer

2/b. The average life of the pendulum is determined by the damping effect, which reduces the amplitude over time. The factor of 2 in the expression 2/b arises from the relationship between the damping force and the exponential decay of amplitude, leading to the time it takes for the amplitude to decrease to 1/e of its initial value.

Question 38

The amplitude of a damped oscillator decreases to 0.9 times its original magnitude in 5s. In another 10s it will decrease to α times its original magnitude, where α equals

Accepted Answer

0.729. The amplitude of a damped oscillator decreases exponentially over time. Given that it decreases to 0.9 of its original value in 5 seconds, we can use the exponential decay formula to find the amplitude after an additional 10 seconds, resulting in a decrease to approximately 0.729 of its original magnitude.

Question 39

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is V0 and its pressure is P0. The piston is slightly displaced from t

Accepted Answer

1/(2π) √(A²γP0/(MV0)). Displacing the piston by x changes gas pressure; for adiabatic isolation the restoring force is F = -(gamma*P0*A^2/V0)*x. So omega = sqrt(gamma*P0*A^2/(M*V0)) and f = (1/2pi) sqrt(A^2*gamma*P0/(M*V0)).

Question 40

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in same direction, then:

Accepted Answer

time period of oscillations is 6τ. The particle's motion is characterized by its displacement over time, and the distances traveled in successive intervals suggest a specific relationship between time and distance in simple harmonic motion. Given that the distances traveled are in a ratio of 1:2, this indicates that the total distance covered in the first two intervals corresponds to a complete cycle, leading to the conclusion that the time period of oscillations is 6τ.

Question 41

A pendulum of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to TM. If the Young's modulus of the material of the wire is Y then 1/Y is equal to: (g = gravitational acceleration)

Accepted Answer

[(TM/T)² - 1] A/(Mg). Adding mass M stretches the wire by dL = MgL/(AY), so T_M^2/T^2 = (L+dL)/L = 1 + Mg/(AY). Hence Mg/(AY) = (T_M/T)^2 - 1, giving 1/Y = [(T_M/T)^2 - 1]*A/(Mg).

Question 42

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is:

Accepted Answer

7A/3. At x = 2A/3, v^2 = w^2(A^2 - 4A^2/9) = 5w^2A^2/9. Tripling speed gives v'^2 = 9*v^2 = 5w^2A^2. New amplitude: 5w^2A^2 = w^2(A'^2 - 4A^2/9) -> A'^2 = 49A^2/9 -> A' = 7A/3.

Question 43

A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 10¹²/sec. What is the force constant of the bonds connecting one atom with the other? (Mole wt. of silver = 108 and Avogadro number = 6.02 × 10²³ gm mole⁻¹)

Accepted Answer

7.1 N/m. Mass of one Ag atom = 108/(6.02e23) g = 1.79e-25 kg. Force constant k = m*omega^2 = m*(2*pi*f)^2 = 1.79e-25*(2*pi*1e12)^2 = 7.1 N/m.

Question 44

A sonometer wire of length 1.5 m is made of steel. The tension in it produces an elastic strain of 1%. What is the fundamental frequency of steel if density and elasticity of steel are 7.7 × 10³ kg/m³ and 2.2 × 10¹¹ N/m² respectively?

Accepted Answer

178.2 Hz. The fundamental frequency of a vibrating wire can be calculated using the formula that incorporates the tension, length, density, and elasticity of the material. Given the parameters for the steel wire and the specified elastic strain, the calculations yield a fundamental frequency of 178.2 Hz, confirming option B as the correct answer.

Question 45

In an oscillating LC circuit the maximum charge on the capacitor is Q. The charge on the capacitor when the energy is stored equally between the electric and magnetic field is

Accepted Answer

Q/√2. At the point where the energy is equally distributed between the electric field of the capacitor and the magnetic field of the inductor, the charge on the capacitor is at a value of Q/√2. This is derived from the relationship between energy stored in the capacitor and inductor, where the total energy remains constant.

Question 46

At t = 0, a capacitor of capacitance C carrying an initial charge q0 is joined to an inductor of self-inductance L. The instant when the energy is shared equally by the electric field and the magnetic field is

Accepted Answer

(π/4)√LC. Capacitor energy ~ q^2 = q0^2*cos^2(wt) with w=1/sqrt(LC). Equal sharing of energy needs cos^2(wt)=1/2, i.e. wt=pi/4, so t = (pi/4)*sqrt(LC).

Question 47

An LCR circuit can be treated as an analogue of a damped spring–mass system. If the mechanical oscillator has mass m, spring constant k, and damping coefficient b, which correspondence between electrical and mechanical quantities is correct?

Accepted Answer

L corresponds to m, C corresponds to k, and R corresponds to b. The damped LCR-oscillator analogy is L corresponds to mass m, R corresponds to damping b, and inverse capacitance 1/C corresponds to spring constant k. Among the choices the one pairing L with m and R with b is correct.

Question 48

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a and in next τ s it travels 2a, in same direction, then -
(1) time period of oscillations is 8τ
(2) amplitude of motion is 4a
(3) time period of oscillations is 6τ

Accepted Answer

(3) time period of oscillations is 6τ. The particle's distances traveled in successive time intervals indicate a quadratic relationship, characteristic of simple harmonic motion. The total distance covered suggests that the time period is 6τ, as the distances align with the expected behavior of a harmonic oscillator.

Question 49

A pendulum made of a uniform wire of cross-sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to Tm. If the Young's modulus of the material of the wire is Y then 1/Y is equal to (g = gravitational acceleration)

Accepted Answer

[(Tm/T)² - 1] A/(Mg). The correct option is derived from the relationship between the time period of a pendulum and the properties of the material, where the change in time period due to the added mass can be expressed in terms of Young's modulus, cross-sectional area, and gravitational force. This option accurately reflects how the increase in mass affects the time period, leading to the derived formula.

Question 50

A particle performs simple harmonic motion with amplitude A. Its speed is trebled at the instant that it is at a distance 2A/3 from equilibrium position. The new amplitude of the motion is:

Accepted Answer

7A/3. When the speed of the particle is increased while it is at a distance of 2A/3 from the equilibrium, the total mechanical energy of the system increases. Since the total energy in simple harmonic motion is proportional to the square of the amplitude, the new amplitude can be calculated using the relationship between the initial and final speeds and positions, leading to the result of 7A/3.

JEE Main Physics: Oscillations questions with solutions

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