Oscillations — NEET Physics PYQ MCQs with Solutions

Solution: For SHM, the restoring force is always opposite to displacement and proportional to it. Hooke's law: $F = -kx$ where $k$ is the spring constant. Here, comparing with $F = -AKx$, we see the force is directly proportional to displacement with a negative sign (indicating restoring nature).

Solution: Given: Extension $x = 5$ cm $= 0.05$ m, Force $F = 10$ N, Mass $m = 2$ kg Spring constant: $k = \frac{F}{x} = \frac{10}{0.05} = 200$ N/m Time period: $T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 2\pi\sqrt{0.01} = 2\pi \times 0.1 = 0.628$ s

Solution: For SHM, acceleration: $a = -\omega^2 x$ Given: $a = 20$ m/s², $x = 5$ m Therefore: $20 = \omega^2 \times 5 \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2$ rad/s Time period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$ s

Solution: Time period of spring-mass system: $T = 2\pi\sqrt{\frac{m}{k}}$ For mass $M$: $T = 2\pi\sqrt{\frac{M}{k}}$ For mass $2M$: $T' = 2\pi\sqrt{\frac{2M}{k}} = \sqrt{2} \cdot 2\pi\sqrt{\frac{M}{k}} = \sqrt{2}T$

Solution: For the mass to just detach from the pan, the spring force must equal the weight at extreme position. At equilibrium: $ka_0 = mg$ For detachment: The mass gets detached when normal force becomes zero, which happens when acceleration is maximum (at extreme position). Restoring force: $F = ka = mg$ (for minimum amplitude) $a = \frac{mg}{k} = \frac{2 \times 10}{200} = \frac{20}{200} = 0.1$ m $= 10$ cm

Solution: Velocity in SHM: $v = \omega\sqrt{a^2 - x^2}$ Case I: $10 = \omega\sqrt{a^2 - 16}$ ... (i) Case II: $8 = \omega\sqrt{a^2 - 25}$ ... (ii) Dividing (ii) by (i): $\frac{8}{10} = \sqrt{\frac{a^2 - 25}{a^2 - 16}}$ Squaring: $\frac{64}{100} = \frac{a^2 - 25}{a^2 - 16}$ Solving: $a^2 = 36.99 \approx 36$ cm² From (i): $\omega = 2$ rad/s Time period: $T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi$ sec

Solution: When a spring is cut in ratio $1:2:3$, the spring constant is inversely proportional to length. If original spring has constant $k$ and total length $L$, then: $k \cdot L = \text{constant}$ For pieces of lengths in ratio $1:2:3$: $k_1 = 6x$, $k_2 = 3x$, $k_3 = 2x$ (where $6x + 3x + 2x = k$ in length ratios) In series: $\frac{1}{k'} = \frac{1}{6x} + \frac{1}{3x} + \frac{1}{2x} = \frac{6}{6x} \Rightarrow k' = x$ In parallel: $k'' = 6x + 3x + 2x = 11x$ Ratio: $\frac{k'}{k''} = \frac{x}{11x} = \frac{1}{11}$

Solution: In simple harmonic motion, both kinetic energy and potential energy attains their maximum value two times in one complete oscillation. Hence, frequency of kinetic energy and potential energy is 2 for one complete oscillation. So, the frequency of the potential energy of a body executing SHM with frequency n is 2n.

Solution: sin t and cos t, both are periodic function of period 2π. We know that, sum of two periodic functions is also a periodic function, hence, sin t + cos t represents periodic motion.

Solution: In SHM, equation of displacement of a particle is y = a sin ωt and equation of acceleration of a particle is A = -aω² sin ωt = aω² sin(ωt + π). Phase difference between displacement and acceleration of a particle is = (ωt + π) - ωt = π rad. Hence, correct option is (d).

Solution: In a simple harmonic motion (SHM) the particle oscillates about its mean position on a straight line. The particle moves from its mean position (O) to an extreme position (P) and then return to its mean position covering same distance of A. Then by the conservative force, it is moved in opposite direction to a point Q by distance A and then back to mean position covering a distance of A. In one time period: x = OP + PO + OQ + QO = A + A + A + A = 4A

Solution: The average velocity of a particle executing simple harmonic motion (SHM) is v_av = (Total displacement)/(Time interval) = (x_f - x_i)/T. In vibrational motion, the particle executes SHM about its mean position. After one complete vibration, the particle reaches its initial position. Therefore, displacement, x_f - x_i = 0. Hence, v_av = 0.

Solution: The displacement of given particle is y = A_0 + A sin ωt + B cos ωt. The general equation of SHM can be given as x = a sin ωt + b cos ωt. From this, we can say that A_0 be the value of mean position. Amplitude, R = √(A² + B² + 2AB cos θ). Since sine and cosine have phase shift of 90°, R = √(A² + B²) [since cos 90° = 0]

Solution: Let O be the centre of circle. At t = 0, the displacement y is maximum and has value 3 m. The general equation of displacement of a particle will be in the form y = A cos ωt. Here, A = 3 m. Then, ω = 2π/T = 2π/4 = π/2. Therefore, y = 3 cos(π/2)t or y = 3 cos(2π/T)t (in metre)

Solution: Magnitude of velocity of particle when it is at displacement x from mean position: |v| = ω√(A² - x²). Magnitude of acceleration: |a| = ω²x. Given, when x = 2 cm, |v| = |a|. So ω√(A² - x²) = ω²x. Therefore ω = √(A² - x²)/x = √(9 - 4)/2 = √5/2. Angular velocity ω = √5/2. Time period T = 2π/ω = 4π/√5 = 4π/√5 · √5/√5 = 4√5π/5 = 4/(5π) s

Solution: Given, $y_1 = a\sin ωt$ and $y_2 = b\cos ωt = b\sin(ωt + π/2)$. The resultant displacement is given by $y = y_1 + y_2 = \sqrt{a^2 + b^2}\sin(ωt + φ)$. Hence, the motion of superimposed wave is simple harmonic with amplitude $\sqrt{a^2 + b^2}$.

Solution: Let A be the amplitude of oscillation. Then $v_1^2 = ω^2(A^2 - x_1^2)$ and $v_2^2 = ω^2(A^2 - x_2^2)$. Subtracting, we get $v_1^2 - v_2^2 = ω^2(x_2^2 - x_1^2)$. Therefore $ω^2 = (v_1^2 - v_2^2)/(x_2^2 - x_1^2)$. Since $T = 2π/ω$, we have $T = 2π\sqrt{(x_2^2 - x_1^2)/(v_1^2 - v_2^2)}$

Solution: For a particle executing SHM, maximum acceleration is α = Aω² and maximum velocity is β = Aω. Dividing α by β, we get α/β = ω. Therefore, ω = α/β. Since T = 2π/ω = 2πβ/α, the time period of vibration is $T = 2πβ/α$.

Solution: As x = A cos ωt, we have v = dx/dt = -Aω sin ωt and a = d²x/dt² = -Aω² cos ωt. At t = 0, a = -Aω². At t = T/4, a = 0. At t = T/2, a = Aω². At t = 3T/4, a = 0. At t = T, a = -Aω². This condition is represented by graph in option (c).

Solution: For simple harmonic motion, acceleration (a) ∝ -displacement (y). The equations y = sin t - cos t and y = 5 cos(3π/4 - 3t) satisfy this condition. The equation y = 1 + t + 2t² is not periodic and y = sin³ t is periodic but not simple harmonic motion.

Solution: For SHM, acceleration (a) ∝ -displacement (x). Given x = a sin² ωt. Differentiating: dx/dt = 2a sin ωt cos ωt = a sin 2ωt. Again differentiating: d²x/dt² = 2aω² cos 2ωt. The given equation does not satisfy the condition for SHM. Therefore, motion is not simple harmonic.

Solution: The condition for a body executing SHM is F = -kx, so a = -ω²x or acceleration ∝ -(displacement). For option (b), let y = x + a, then acceleration = -k(x + a) = -ky, which represents SHM about the position y = 0 (or x = -a).

Solution: Maximum acceleration of body executing SHM is given by a_max = ω²a. For two different cases: a_max1/a_max2 = (ω₁²a)/(ω₂²a) = (ω₁/ω₂)² = (100/1000)² = (1/10)² = 1/100 = 1:100

Solution: Equation of motion is x = a sin(ωt + π/6). Velocity v = dx/dt = aω cos(ωt + π/6). Maximum velocity is v_max = aω. When v = v_max/2: aω/2 = aω cos(ωt + π/6), so cos(ωt + π/6) = 1/2. This gives ωt + π/6 = π/3, so ωt = π/6. Since ω = 2π/T, we have t = T/12.

Solution: In simple harmonic motion, the total energy is constant. When KE = K₀ cos² ωt, the maximum kinetic energy is K₀. At this point (mean position), PE = 0. At extreme positions, KE = 0 and PE = K₀ (maximum). Total energy = K₀. Therefore, maximum PE = K₀ and total energy = K₀.

Solution: Let displacement equation be x = a sin ωt. When particle travels half of amplitude from equilibrium: x = a/2. So a/2 = a sin ωt, which gives sin ωt = 1/2 = sin(π/6). Therefore ωt = π/6. Since ω = 2π/T, we have t = T/12.

Solution: Maximum speed of a particle executing SHM is given by v_max = aω = a(2πn), where n is frequency. Therefore n = v_max/(2πa) = 31.4/(2 × 3.14 × 5) = 31.4/31.4 = 1 Hz

Solution: For SHM with displacement x = a sin ωt: Velocity v = aω cos ωt = ω√(a² - x²), and acceleration a = -ω²x. When x = 0 (equilibrium), v = aω = v_max and a = 0. When x = ±a (extreme positions), v = 0 and a = ∓ω²a = a_max. Thus when v is maximum, a is minimum (zero).

Solution: Potential energy U = (1/2)mω²x². When particle is halfway to end point, x = a/2. So U = (1/2)mω²(a/2)² = (1/4)[(1/2)mω²a²] = E/4, where E is total energy.

Solution: Potential energy is given by U = (1/2)kx². The graph is parabolic. At equilibrium position (x = 0), potential energy is minimum. At extreme positions x₁ and x₂, potential energies are U₁ = (1/2)kx₁² and U₂ = (1/2)kx₂² respectively. Kinetic energy is maximum at mean position and zero at extreme positions.