Oscillations practice · Medium

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.

The correct answer is D. The force constant is inversely proportional to the length of the spring, so when the spring is cut, the force constants change accordingly.

Option B (Inertia of the system) is correct because the force constant $k$ depends on the elasticity of the system, extension/displacement, and air resistance, but not on the inertia of the system.

Option A ($0.15 \text{{ m}}$) is correct because it is calculated by equating the kinetic energy of the mass to the potential energy stored in the spring.

Option C (maximum) is correct because in simple harmonic motion, the velocity is maximum at the mean or equilibrium position.

Option A ($\sqrt{2} f$) is correct because cutting the spring in half doubles the spring constant, which increases the frequency by a factor of $\sqrt{2}$.

Option A ($10 \sqrt{3}$) is correct because it is calculated using the SHM velocity formula $v = \omega \sqrt{A^2 - y^2}$ where $y$ is the distance from the mean position.

Option B ($a / \sqrt{2}$) is correct because at this displacement, the kinetic and potential energies are equal in simple harmonic motion.

Option D ($T = 2\pi \sqrt {\frac{{x_2^2 - x_1^2}}{{v_1^2 - v_2^2}}}$) is correct because it correctly relates the time period to the given velocities and positions in SHM.

The correct answer is B because both the assertion and reason are true, but the reason is not the correct explanation of the assertion. The time period depends on the spring constant, but the explanation focuses on the relationship between hardness and the spring constant.

Option D ($36.28 \mathrm{~s}$) is correct because it calculates the time lost by the pendulum clock due to the increase in temperature, considering the linear expansion of the steel wire.

Option C ($\frac{5 \pi}{3}$) is correct because it satisfies the given conditions of position and velocity at t=0.

Option A is correct because oscillations in a spring-mass system involve the continuous exchange of energy between the block and the spring, and the total energy of the system remains constant (assuming no damping).

Option A (If both assertion and reason are true and reason is the correct explanation of the assertion) is correct because SHM requires acceleration to be proportional and opposite in direction to displacement.

Option B is correct because $\sin^2(\omega t)$ is a periodic function but not a simple harmonic motion, and its period is $\pi/\omega$.

Option B ($T_2>T_3>T_1$) is correct because the acceleration due to gravity decreases with altitude and depth, but the decrease is greater at altitude h than at depth h.

Option D (4) is correct because the percentage of potential energy is calculated by dividing the square of the displacement by the square of the amplitude and multiplying by 100.

Option D ($8 \sqrt{2} \mathrm{~m}$) is correct because the amplitude of oscillation is calculated using the given velocity, time period, and initial conditions of the SHM.

Option A (2 Hz) is correct because the frequency of kinetic energy in SHM is twice the frequency of the displacement.

Option C ($T = 2\pi \sqrt {\left( {\frac{{m\left( {{{K_1} + {K_2}}}\right)}}{{{{K_1}{K_2}}}}} \right)}$) is correct because it uses the equivalent spring constant for springs in series to calculate the time period.

Option D (S-I is false & S-II is true) is correct because the total mechanical energy in SHM is constant, while both kinetic and potential energies oscillate with twice the frequency of the SHM.

Option C (First increase and then decrease to the original value) is correct because the center of mass initially lowers, increasing the effective length, and then returns to the original position as the water empties.