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Simple harmonic motion and spring oscillations: NEET MCQs with answers

75 questions on this topic of Oscillations and Waves (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. The total mechanical energy of a particle executing simple harmonic motion is E. When its displacement from the mean position is half of its amplitude, the kinetic energy of the particle is:

  1. E / 4
  2. E / 2
  3. E / √2
  4. 3E / 4
Show answer and explanation

Answer: (D) 3E / 4

Potential energy at displacement x is U = (1/2) k x². For x = A/2, U = (1/2) k (A/2)² = (1/4) [(1/2) k A²] = E / 4. Therefore, kinetic energy is K = E - U = E - E/4 = 3E / 4.

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2. A block of mass 1 kg is connected to a horizontal spring of spring constant 100 N/m. The other end of the spring is fixed. The block is pulled a distance of 0.1 m from equilibrium on a frictionless surface and released from rest at t = 0. The time taken to first reach a displacement of 0.05 m is:

  1. π / 60 s
  2. π / 30 s
  3. π / 20 s
  4. π / 15 s
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Answer: (B) π / 30 s

The angular frequency is ω = √(k/m) = √(100/1) = 10 rad/s. Since the block starts from rest at maximum displacement, x(t) = A cos(ωt). We set 0.05 = 0.1 cos(10t), which gives cos(10t) = 1/2. Thus, 10t = π/3, yielding t = π / 30 s.

hard

3. A uniform spring of stiffness k is cut into two parts whose lengths are in the ratio 2 : 3. What is the equivalent spring constant when these two parts are connected in parallel?

  1. 5k/6
  2. 15k/4
  3. 6k/5
  4. 25k/6
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Answer: (D) 25k/6

Spring stiffness is inversely proportional to length (k ∝ 1/L). If the lengths are L₁ = (2/5)L and L₂ = (3/5)L, then k₁ = (5/2)k and k₂ = (5/3)k. When connected in parallel, the equivalent spring constant is k_eq = k₁ + k₂ = (5/2)k + (5/3)k = 25k/6.

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4. Two simple harmonic motions are represented by y₁ = 4 sin(ωt + π/6) and y₂ = 3 cos(ωt). The amplitude of the resultant motion obtained by their superposition is:

  1. 5
  2. √13
  3. √37
  4. 7
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Answer: (C) √37

We rewrite y₂ as y₂ = 3 sin(ωt + π/2). The phase difference between the two motions is Δϕ = π/2 - π/6 = π/3 (60°). The resultant amplitude is A = √(A₁² + A₂² + 2 A₁ A₂ cos Δϕ) = √(4² + 3² + 2 × 4 × 3 × cos 60°) = √(16 + 9 + 24 × 0.5) = √(25 + 12) = √37.

easy

5. The potential energy of a particle of mass 0.1 kg executing simple harmonic motion is given by U(x) = 20x² + 5, where U is in joules and x is displacement from the origin in metres. What is the frequency of oscillation of the particle?

  1. (2.5/π) Hz
  2. (5/π) Hz
  3. (20/π) Hz
  4. (10/π) Hz
Show answer and explanation

Answer: (D) (10/π) Hz

The potential energy is U(x) = 20x² + 5. The restoring force is F = -dU/dx = -40x. Thus, the effective spring constant is k = 40 N/m. The angular frequency is ω = √(k/m) = √(40/0.1) = √400 = 20 rad/s. Therefore, the frequency is f = ω / (2π) = 20 / (2π) = 10/π Hz.

hard

6. A block of mass m rests on a smooth horizontal surface and is connected to a fixed wall via a spring of force constant k. A constant horizontal pulling force F is suddenly applied to the block directed away from the wall. If the spring was initially in its natural unstretched length, what is the maximum speed attained by the block during the subsequent motion?

  1. F√(2 / mk)
  2. 2F / √(mk)
  3. F / (2√(mk))
  4. F / √(mk)
Show answer and explanation

Answer: (D) F / √(mk)

Under the constant force F, the new equilibrium position (mean position) is at an extension x₀ where k x₀ = F, so x₀ = F/k. Since the block was released from rest at x = 0 (the unstretched position), the amplitude of oscillation about the new equilibrium position is A = x₀ = F/k. The maximum speed during SHM occurs at the mean position and is given by v_max = ω A = √(k/m) · (F/k) = F / √(mk).

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7. A particle executes SHM of amplitude A and time period T. Starting from the positive extreme position, what is the minimum time taken by the particle to reach a displacement of -A/2?

  1. T / 6
  2. T / 4
  3. 5T / 12
  4. T / 3
Show answer and explanation

Answer: (D) T / 3

Let displacement be x(t) = A cos(ωt). At the initial extreme position t = 0, x = A. When x = -A/2, cos(ωt) = -1/2, giving the smallest positive phase ωt = 2π/3. Since ω = 2π/T, we have (2π/T)t = 2π/3, which gives t = T/3.

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8. A block of mass m is suspended from two identical springs, each of spring constant k, connected in parallel. If this system oscillates with a time period T₁, and when the same two springs are connected in series with the same mass it oscillates with a time period T₂, the ratio T₁/T₂ is:

  1. 1 : 4
  2. 4 : 1
  3. 2 : 1
  4. 1 : 2
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Answer: (D) 1 : 2

For two identical springs in parallel, k_p = 2k, so T₁ = 2π√(m / 2k). For series connection, k_s = k/2, so T₂ = 2π√(m / (k/2)) = 2π√(2m / k). The ratio T₁/T₂ = √[(m/2k) / (2m/k)] = √(1/4) = 1/2.

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9. A 0.25 kg mass attached to a spring of force constant 20 N/m oscillates horizontally. Its time period is:

  1. 0.702 s
  2. 0.351 s
  3. 56.2 s
  4. 0.112 s
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Answer: (A) 0.702 s

For a spring-mass oscillator T = 2π√(m/k) = 2π√(0.25/20) = 0.702 s.

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10. A 0.1 kg mass attached to a spring of force constant 60 N/m oscillates horizontally. Its time period is:

  1. 0.0408 s
  2. 0.257 s
  3. 0.128 s
  4. 154 s
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Answer: (B) 0.257 s

For a spring-mass oscillator T = 2π√(m/k) = 2π√(0.1/60) = 0.257 s.

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11. A block of mass M is attached to a horizontal ideal spring of spring constant k and executes SHM with an amplitude A on a frictionless surface. As it passes through its mean position, a small blob of putty of mass m is gently dropped on it and sticks to it. The new amplitude of oscillation becomes:

  1. A √((M + m) / M)
  2. A √(M / (M + m))
  3. A (M / (M + m))
  4. A ((M + m) / M)
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Answer: (B) A √(M / (M + m))

At the mean position, the speed of M is v₀ = ωA = A√(k/M). By conservation of linear momentum, M v₀ = (M + m)v', giving v' = [M/(M + m)] A√(k/M). The new maximum speed is also related to the new amplitude A' by v' = ω' A' = A'√(k/(M + m)). Equating these: A'√(k/(M + m)) = [M/(M + m)] A√(k/M) ⇒ A' = A √(M / (M + m)).

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12. A 0.1 kg mass attached to a spring of force constant 120 N/m oscillates horizontally. Its time period is:

  1. 0.181 s
  2. 218 s
  3. 0.0289 s
  4. 0.0907 s
Show answer and explanation

Answer: (A) 0.181 s

For a spring-mass oscillator T = 2π√(m/k) = 2π√(0.1/120) = 0.181 s.

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