Oscillations practice · Mixed difficulty

The path of the particle is an ellipse (Option B) because the x and y coordinates satisfy the equation of an ellipse after integration and elimination of the time variable.

Option B (its time period is infinite at that point) is correct because the time period becomes infinite when the distance from the center of gravity is zero.

Option B (Restoring force) is correct because oscillatory motion requires a force that pulls the object back towards its equilibrium position.

Option B (10) is correct because the two pendulums will be in phase again after the longer pendulum completes 10 oscillations, as calculated by equating the time taken for a certain number of oscillations.

Option D (432 s) is correct because the time lost or gained is directly proportional to the percentage change in the length of the pendulum.

Option A is correct because the acceleration in SHM is greatest when the spring is at maximum displacement (fully compressed or extended) due to Hooke's Law.

Option C (1.6 kg) is correct because the mass is calculated using the period of oscillation formula for a spring-mass system.

Option A (${\pi ^2}m{a^2}{\mathrm{v}^2}$) is correct because the average kinetic energy is calculated as $\frac{1}{4}m a^2 \omega^2$, where $\omega = 2 \pi v$.

Option A (S-I & S-II are true) is correct because oscillatory motions are periodic, but not all periodic motions are oscillatory, and a simple pendulum exemplifies oscillatory motion.

Option C is correct because the oscillating particle and the driving force can have different phases, but this phase difference depends on the driving frequency relative to the natural frequency, not the initial conditions.

Option A (The distance of particle from its mean position) is correct because displacement in oscillatory motion is defined as the distance from the mean position.

Option A is correct because in SHM, velocity leads displacement by π/2, and acceleration leads velocity by π/2, making the phase difference between velocity and acceleration π/2.

Option C (More than $T$) is correct because the acceleration due to gravity on the moon is less than that on Earth, leading to a longer period.

Option A ($\mathrm{\pi }\,\,\mathrm{rad}$) is correct because acceleration in SHM is always opposite in direction to the displacement.

Option B (Its kinetic energy decreases while potential energy increases) is correct because as the swing rises, its speed decreases (decreasing KE) and its height increases (increasing PE).

Option C (Driving frequency is the same as the natural frequency of the oscillator) is correct because resonance occurs when the driving frequency matches the natural frequency.

Option D (Same at all position) is correct because in simple harmonic motion, the total energy remains constant, interchanging between kinetic and potential forms.

Option A (S-I & S-II are true) is correct because both statements accurately describe the behavior of damped oscillations.

Option B ($\infty$) is correct because in a freely falling cabin, the effective acceleration due to gravity is zero, leading to an infinite time period for the pendulum.

Option C (an ellipse) is correct because the relationship between velocity and displacement in SHM forms an ellipse equation.

Option B (If both assertion and reason are true but reason is not the correct explanation of assertion) is correct because the Earth's revolution is periodic but not oscillatory, and the reason correctly states the relationship between oscillatory and periodic motion, but it doesn't explain why the Earth's motion is periodic.

Option B (2f) is correct because the potential energy in SHM oscillates at twice the frequency of the displacement.

Option A ($2 \mathrm{~sec}$) is correct because the time period of a simple pendulum is independent of the mass of the bob.

Option D (If both Assertion & Reason are false) is correct because SHM is a periodic motion, and periodic motion repeats its position after a certain interval.

Option A ((A), (R) are true : (R) is the correct explanation of (A)) is correct because the spring constant is inversely proportional to the length of the spring, so cutting it in half doubles the spring constant of each piece.

Option A (a) $\frac{A_2 \sqrt{3}}{2};$ (b) $\omega \sqrt{A_1^2+A_2^2+A_1 A_2}$ is correct because it correctly calculates the displacement at t=0 and the maximum speed of the particle.

The correct answer is periodic but not simple harmonic because the given equation can be expanded into a series of sine functions with different frequencies, indicating a periodic but non-SHM motion.

Option D is correct because the time period of a spring-mass system depends only on the mass and spring constant, not on the gravitational acceleration, so it remains the same on the Moon.

Option D ($2 \sqrt{2} \mathrm{~cm}$) is correct because it is the displacement at which the potential and kinetic energies are equal in SHM.

Option C (Periodic but not simple harmonic) is correct because uniform circular motion is periodic, but its projection onto a diameter is simple harmonic.