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Kinetic and potential energy in SHM: NEET MCQs with answers

50 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 potential energy of a particle of mass 0.25 kg executing SHM is zero at the mean position. If its speed while passing through the mean position is 4 m/s, the total mechanical energy of the oscillator is:

  1. 1.0 J
  2. 4.0 J
  3. 0.5 J
  4. 2.0 J
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Answer: (D) 2.0 J

At the mean position, the potential energy is zero, so the total mechanical energy equals the maximum kinetic energy. Therefore, E = (1/2)mv_max² = (1/2) × 0.25 × 4² = 0.5 × 0.25 × 16 = 2.0 J.

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2. The potential energy function of a 0.2 kg particle moving in a straight line is given by U(x) = (20x² - 40x + 30) J, where x is in meters. If the total mechanical energy of the particle is 25 J, the maximum kinetic energy attained during oscillation is:

  1. 10 J
  2. 20 J
  3. 5 J
  4. 15 J
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Answer: (D) 15 J

Rewriting U(x) in standard form: U(x) = 20(x² - 2x + 1) + 10 = 20(x - 1)² + 10 J. The minimum potential energy is U_min = 10 J at the equilibrium position x = 1 m. Since total energy E = K + U = 25 J, the maximum kinetic energy is K_max = E - U_min = 25 - 10 = 15 J.

hard

3. The potential energy of a particle of mass 1 kg moving along the x-axis is given by U(x) = 50(x - 0.2)² + 10, where U is in joules and x is in metres. If the total mechanical energy of the particle is 35 J, the maximum speed of the particle is:

  1. 5 m/s
  2. 2√5 m/s
  3. 10 m/s
  4. 5√2 m/s
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Answer: (D) 5√2 m/s

The minimum potential energy occurs at x = 0.2 m and is U_min = 10 J. Maximum kinetic energy occurs where potential energy is minimum: K_max = E - U_min = 35 - 10 = 25 J. Since K_max = (1/2)mv_max², we have (1/2)(1)v_max² = 25, giving v_max² = 50, hence v_max = √50 = 5√2 m/s.

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4. A particle of mass 0.4 kg executes simple harmonic motion along the x-axis. Its total mechanical energy is 1.8 J, and the potential energy at the mean position is zero. If the amplitude of oscillation is 0.3 m, the maximum acceleration of the particle is:

  1. 20 m/s²
  2. 15 m/s²
  3. 30 m/s²
  4. 45 m/s²
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Answer: (C) 30 m/s²

The total energy is E = (1/2)mω²A². Substituting the values: 1.8 = (1/2)(0.4)ω²(0.3)², which gives 1.8 = 0.2 × 0.09 × ω² = 0.018 ω², so ω² = 100 rad²/s² and ω = 10 rad/s. The maximum acceleration is a_max = ω²A = 100 × 0.3 = 30 m/s².

easy

5. A particle of mass m executes SHM along the x-axis with an amplitude A and angular frequency ω. If the potential energy is taken to be zero at the mean position, what is the ratio of kinetic energy to potential energy when the displacement is x = A/√5?

  1. 1 : 4
  2. 4 : 1
  3. 5 : 1
  4. 2 : 1
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Answer: (B) 4 : 1

Kinetic energy is K = (1/2)k(A² - x²) and potential energy is U = (1/2)kx². Substituting x = A/√5 gives K = (1/2)k(A² - A²/5) = (4/5)(1/2)kA² and U = (1/5)(1/2)kA². Thus, the ratio K/U = 4/1.

hard

6. A particle of mass m executes SHM about x = 0 with amplitude A. When passing through x = A/2, an external impulse suddenly doubles its speed without changing its direction or position. The new amplitude of oscillation becomes:

  1. (√7/2) A
  2. (√10/2) A
  3. (√13/2) A
  4. 2 A
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Answer: (C) (√13/2) A

Initially at x = A/2, the speed is v = ω√(A² - x²) = ω√(3/4 A²) = (√3/2)ωA. The potential energy is U = (1/2)mω²(A/2)² = (1/8)mω²A². The new speed is v' = 2v = √3 ωA. The new kinetic energy is K' = (1/2)m(v')² = (3/2)mω²A². The new total energy is E' = U + K' = (1/8 + 3/2)mω²A² = (13/8)mω²A². Since E' = (1/2)mω²(A')², we get (A')² = 13A²/4, so A' = (√13/2)A.

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7. The potential energy of a particle of mass 250 g executing SHM along the x-axis is given by U = 20 + 80(x - 0.1)², where U is in joules and x is in metres. If the particle oscillates with an amplitude of 0.05 m, what is its maximum kinetic energy?

  1. 0.10 J
  2. 20.20 J
  3. 0.20 J
  4. 0.05 J
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Answer: (C) 0.20 J

The potential energy is U = U₀ + (1/2)k(x - x₀)². Comparing gives k/2 = 80, so k = 160 N/m. The maximum kinetic energy corresponds to (1/2)kA² = (1/2) × 160 × (0.05)² = 80 × 0.0025 = 0.20 J.

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8. A body executing SHM has maximum speed v_max and maximum acceleration a_max. The total mechanical energy of the oscillator of mass m is:

  1. m(v_max)² / 2
  2. m(v_max)⁴ / (2 a_max²)
  3. m(a_max)² / (2 v_max²)
  4. m(v_max)³ / (2 a_max)
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Answer: (A) m(v_max)² / 2

In SHM, the maximum speed occurs at the mean position where the potential energy is minimum (taken as zero). Therefore, the total mechanical energy is entirely kinetic at this point, giving E = (1/2)m(v_max)².

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9. Consider the following statements regarding a particle executing SHM about its mean position: Statement I: The average kinetic energy over one complete cycle equals the average potential energy over one complete cycle (assuming U = 0 at x = 0). Statement II: The average kinetic energy over one complete cycle equals half of the total mechanical energy. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
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Answer: (A) Both Statement I and Statement II are correct

The time average of sin²(ωt) and cos²(ωt) over a full cycle is 1/2. Therefore, ⟨K⟩ = (1/4)kA² and ⟨U⟩ = (1/4)kA², meaning ⟨K⟩ = ⟨U⟩ = E/2. Both statements are correct.

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10. A particle of mass m executes SHM about x = 0. Its kinetic energy K varies with position x according to K = α - βx², where α and β are positive constants. The frequency of oscillation of the particle is:

  1. (1/2π) √(2β/m)
  2. (1/2π) √(β/m)
  3. (1/2π) √(α/m)
  4. (1/π) √(β/2m)
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Answer: (A) (1/2π) √(2β/m)

In SHM, kinetic energy is K = (1/2)mω²A² - (1/2)mω²x². Comparing this with K = α - βx² gives (1/2)mω² = β, so ω = √(2β/m). Therefore, the frequency is f = ω/(2π) = (1/2π) √(2β/m).

easy

11. A linear harmonic oscillator has a total mechanical energy of 100 J. When its kinetic energy is 64 J, its potential energy is 36 J. What fraction of its amplitude is its displacement at this instant? (Assume U = 0 at the mean position)

  1. 0.64
  2. 0.8
  3. 0.36
  4. 0.6
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Answer: (D) 0.6

Potential energy is U = (1/2)kx² and total energy is E = (1/2)kA². Thus, U/E = x²/A². Given U = 36 J and E = 100 J, we have x²/A² = 36/100 = 0.36. Taking the square root gives x/A = 0.6.

easy

12. A particle executing SHM has a total mechanical energy E (with U = 0 at the mean position). If its displacement from the mean position is increased by 20% from an initial position x, its potential energy increases by:

  1. 44%
  2. 40%
  3. 20%
  4. 24%
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Answer: (A) 44%

Potential energy in SHM is proportional to the square of displacement: U ∝ x². If the displacement becomes x' = 1.20 x, the new potential energy is U' = (1.20)² U = 1.44 U. The percentage increase is (U' - U)/U × 100% = 44%.

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