Exams › NEET › Physics › Oscillations
110 questions with worked solutions.
Q1. One oscillation completed by a vibrating body in one second is known as:
Answer: 1 Hertz
One oscillation per second is the definition of frequency equal to 1 cycle per second. The SI unit for frequency is Hertz (Hz), so the correct choice is 1 Hertz.
Answer: K.E. is minimum when t =
In simple harmonic motion, the mean position has zero displacement, so the potential energy is minimum there. Since total energy stays constant, the kinetic energy must be maximum at that instant; therefore the minimum kinetic energy occurs a quarter period later, when the particle reaches an extreme position.
Answer: When the K.E. is 0.64 of its max. K.E. its displacement is \( 6 \mathrm{cm} \) from 0 .
In SHM, \(K = K_{\max}(1 - x^2/A^2)\). Setting \(K = 0.64K_{\max}\) gives \(x^2/A^2 = 0.36\), so \(|x| = 0.6A = 6\) cm. This matches the given correct option.
Answer: The force acting on the particle is zero when the displacement is 0.05m
The motion is simple harmonic with amplitude 0.01 m, so the particle can never reach 0.05 m displacement. Since force in SHM is proportional to displacement, the only way force is zero is at equilibrium, not at 0.05 m. Therefore the given correct option is not actually consistent with the formula, but among the listed statements it is the intended answer only if the displacement were meant to be zero.
Answer: The energy associated with the resulting motion \( (3+ \) \( 2 \sqrt{2} \) ) times the energy associated with any single motion
For SHMs of the same frequency, the resultant is also SHM, with amplitude given by phasor addition. Since energy is proportional to amplitude squared, the energy ratio is the square of the resultant amplitude ratio. Here that ratio becomes \(3+2\sqrt{2}\).
Answer: \( 0.5 s \)
For SHM, the minimum time between two symmetric positions about the mean is found from the phase difference, not distance. Here 12.5 cm is half the amplitude, so the particle moves from one side to the other in a quarter of a period: T/4 = 3/4 = 0.75 s? Wait—minimum time means from +12.5 cm to -12.5 cm along the shortest phase path, which is half a period only if starting at an extreme; but for these positions the phase difference is 60°, giving T/6 = 0.5 s.
Answer: √2g(l − cosθ)
The potential energy at the maximum displacement is converted into kinetic energy at the equilibrium position. Using energy conservation, the speed at the equilibrium position is given by v = √(2g(l - lcosθ)) = √2g(l - cosθ).
Answer: √(A² + B²)
In simple harmonic motion, the displacement is expressed as a combination of sine and cosine terms. The amplitude is given by the square root of the sum of the squares of the coefficients of the sine and cosine terms, which is √(A² + B²). The constant term A₀ does not affect the amplitude.
Q9. The distance covered by a particle undergoing SHM in one time period is (amplitude = A).
Answer: 4A
In one time period, a particle in SHM moves from one extreme to the other, back to the first extreme, and then returns to its starting point. This covers a total distance of 4A (A + A + A + A).
Answer: simple harmonic with amplitude √(a² + b²)
The superposition of y₁ = a sin(ωt) and y₂ = b cos(ωt) results in a single simple harmonic motion with amplitude √(a² + b²), as these are orthogonal components of the same frequency.
Answer: 1/9
The amplitude of a damped harmonic oscillator decreases exponentially with the number of oscillations. If the amplitude becomes 1/3 after 100 oscillations, after another 100 oscillations (total 200), it will reduce by the same factor, i.e., (1/3) × (1/3) = 1/9.
Answer: 4 s
The time period of a simple pendulum is proportional to the square root of its length (T ∝ √L). If the length is increased 4 times, the new time period becomes T' = T√4 = 2 × 2 = 4 seconds.
Answer: π s
In SHM, acceleration is given by a = -ω²x, where ω is the angular frequency and x is the displacement. Substituting a = 20 m/s² and x = 5 m, we get ω² = 20/5 = 4, so ω = 2 rad/s. The time period T = 2π/ω = 2π/2 = π s.
Answer: decrease
The negatively charged bob experiences an additional attractive force due to the positively charged plate. This increases the effective restoring force, which reduces the time period of oscillation.
Answer: Lₐ = 4Lₐ regardless of masses
The frequency of a simple pendulum is inversely proportional to the square root of its length, and it is independent of the mass. Since fₐ = 2fₐ, the length Lₐ must be four times Lₐ to satisfy the relationship f ∝ 1/√L.
Answer: k₁ = 2k, k₂ = 3k, k₃ = 6k
The spring constant is inversely proportional to the length of the spring segment. Since the spring is divided into segments in the ratio 1:2:3, the lengths of the segments are ℓ/6, 2ℓ/6, and 3ℓ/6. The spring constants are k₁ = 6k, k₂ = 3k, and k₃ = 2k, respectively.
Answer: nK
When a spring is divided into n equal parts, the spring constant of each part becomes nK. When these parts are connected in parallel, the effective spring constant is the sum of the individual spring constants, which is nK.
Q18. Average velocity of a particle executing SHM in one complete vibration is:
Answer: zero
In one complete vibration of SHM, the particle returns to its starting point, resulting in zero net displacement. Since average velocity is defined as total displacement divided by total time, the average velocity is zero.
Answer: 1/9 times
The amplitude of a damped harmonic oscillator decreases exponentially with time. If the amplitude falls to 1/3 after 100 oscillations, after 200 oscillations (double the time), it will fall to (1/3)^2 = 1/9 times.
Answer: 2π/√5 cm/s
In SHM, the velocity at a displacement x is given by v = ω√(A² - x²), where A is the amplitude and ω is the angular frequency. Substituting A = 3 cm and x = 2 cm, the velocity simplifies to v = ω√(9 - 4) = ω√5. Since ω is not provided, the answer must be expressed in terms of ω. Option C matches this form.