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Free Waves MCQs with Answers

1,281 Waves MCQs from Physics, each with the correct answer and a written explanation of why it is correct. Free and unlimited, with no account needed.

Progressive waves transfer energy through oscillations, with wavelength, frequency, amplitude and wave speed related by v = fλ; sound speed depends on the medium. Superposition produces interference and stationary waves in strings and organ pipes, while simple harmonic motion describes the oscillation itself and must be distinguished from wave propagation.

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Progressive WavesWave SpeedSpeed of SoundSuperposition and InterferenceStationary WavesOrgan PipesSimple Harmonic Motion
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1,281 questions · page 12 of 65

221. A spring of spring constant k is cut into two equal parts. What is the spring constant of each part?

  • A. k/2
  • B. k
  • C. 2k
  • D. 4k

Explanation: The spring constant k is inversely proportional to the length. Cutting the spring in half doubles the spring constant, so each part has…

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Correct answer: 2k

222. A particle in SHM has displacement x(t) = 0.03sin(6t + π/3) (meters). What is the maximum acceleration?

  • A. 0.54 m/s
  • B. 0.72 m/s²
  • C. 1.08 m/s²
  • D. 1.44 m/s2

Explanation: The maximum acceleration amax = Aω² = 0.03 × 62 = 1.08 m/s².

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Correct answer: 1.08 m/s²

223. A block of mass 0.8 kg is attached to a horizontal spring (k = 320 N/m). The block is pulled 0.1 m from equilibrium and released. How long does it take to reach the equilibrium position for the first time?

  • A. 0.0785s
  • B. 0.1 s
  • C. 0.157s
  • D. 0.2 s

Explanation: The angular frequency ω = √k/m = √320/0.8 = 20rad/s. The time to reach equilibrium from maximum displacement is ;t = T/4 = π/2ω = π/2 x 20…

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Correct answer: 0.0785s

224. A mass-spring system is driven at a frequency f = 3 Hz . The natural frequency is f = 2 Hz. What is the ratio of the amplitude at resonance to the amplitude at f (assume light damping)?

  • A. 1.0
  • B. 1.5
  • C. 2.0
  • D. 3.0

Explanation: The amplitude ratio is given by Aresonance / A(fd) = fo / |fd-fo| , 2/ |3-2| = 2.

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Correct answer: 2.0

225. A particle executes SHM with amplitude 0.15 m. When its displacement is 0.05 m, its speed is 0.6 m/s. What is the angular frequency?

  • A. 3.0 rad/s
  • B. 4.47 rad/s
  • C. 5.0 rad/s
  • D. 6.0 rad/s

Explanation: Using v = ωV √A2 - x² , we get 0.6 = ω√(0.15)2 - (0.05)2.Solving, ω = 0.6/√0.02 ≈ 4.47 rad/s.

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Correct answer: 4.47 rad/s

226. Two identical springs (k = 150 N/m) are connected in series, and a 0.6 kg mass is attached. What is the period of oscillation?

  • A. 0.4 s
  • B. 0.562 s
  • C. 0.8 s
  • D. 1.0 s

Explanation: The equivalent spring constant for series springs is keq = k/2 = 150/2 = 75 N/m. The period T = 2π √m/ keq = 2π √0.6/75 ≈ 0.562 s.

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Correct answer: 0.562 s

227. A mass-spring system (m = 0.25 kg, k = 100 N/m) is set into SHM with amplitude 0.1 m. What is the speed when the displacement is half the amplitude?

  • A. 1.0 m/s
  • B. 1.5 m/s
  • C. 1.73 m/s
  • D. 2.0 m/s

Explanation: Using energy conservation, v = ω √A2 - x2 = √k/m √A2 - (A/2)2 = √100/0.25 √0.1 which equals 1.73 m/s.

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Correct answer: 1.73 m/s

228. A simple pendulum of length 0.8 m has a period of 1.5 s on an unknown planet. What is the gravitational acceleration on that planet?

  • A. 9.8 m/s²
  • B. 12.0 m/s²
  • C. 14.05 m/s²
  • D. 16.0 m/s2

Explanation: Using T = 2π √L/g , we get g = 4π2L / T2 = 4π2 (0.8) / (1.5)2 ≈ 14.05 m/s2 .

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Correct answer: 14.05 m/s²

229. A block of mass 0.4 kg is attached to a spring (k = 160 N/m). It is displaced 0.06 m and released. What is the maximum acceleration?

  • A. 12 m/s²
  • B. 18 m/s²
  • C. 24 m/s2
  • D. 30 m/s²

Explanation: The maximum acceleration amax = ω²A = k/m A = 160/0.4(0.06) = 24 m/s2 .

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Correct answer: 24 m/s2

230. A particle in SHM has displacement x(t) = 0.05cos(4πt + π/6) (meters). What is the period of the motion?

  • A. 0.25 s
  • B. 0.5 s
  • C. 1 s
  • D. 2 s

Explanation: The angular frequency ω = 4π. The period T= 2π/ω = 2π/4π = 0.5s.

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Correct answer: 0.5 s

231. A spring is stretched by 0.04 m with a force of 8 N. What is the spring constant?

  • A. 100 N/m
  • B. 150 N/m
  • C. 200 N/m
  • D. 250 N/m

Explanation: The spring constant k = F/x = 8 / 0.04 = 200 N/m.

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Correct answer: 200 N/m

232. A 0.3 kg mass oscillates with amplitude 0.08 m. If its maximum kinetic energy is 0.096 J, what is the angular frequency?

  • A. 5 rad/s
  • B. 8 rad/s
  • C. 10 rad/s
  • D. 12 rad/s

Explanation: The maximum kinetic energy K = ½ mω²A². Substituting ; 0.096 = ½ (0.3)ω2 (0.08)2 . Solving gives ω = 10 rad/s.

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Correct answer: 10 rad/s

233. For a mass-spring system with m = 0.2 kg and k = 80 N/m, what is the frequency of oscillation?

  • A. 1.59 Hz
  • B. 3.18 Hz
  • C. 4.77 Hz
  • D. 6.37 Hz

Explanation: The frequency f is given by f = 1/2π √k/m = 1/2π √80/0.2 ≈ 3.18 Hz.

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Correct answer: 3.18 Hz

234. Centripetal acceleration is also called:

  • A. Angular acceleration
  • B. Tangential acceleration
  • C. Linear acceleration
  • D. Radial acceleration

Explanation: Centripetal acceleration is the acceleration that keeps an object moving in a circular path; it points toward the center of the circle.

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Correct answer: Radial acceleration

235. A particle in SHM has an amplitude of 0.12 m and a maximum speed of 0.6 m/s. What is its angular frequency?

  • A. 3 rad/s
  • B. 5 rad/s
  • C. 6 rad/s
  • D. 8 rad/s

Explanation: The maximum speed vmax in SHM is given by vmax = Aω. Thus, ω = vmax/A = 0.6 / 0.12 = 5 rad/s.

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Correct answer: 5 rad/s

236. A simple pendulum of length 1.6 m is on Earth (g = 9.8m/s²). What is its period of oscillation?

  • A. 2.0 s
  • B. 2.54 s
  • C. 3.0 s
  • D. 4.0 s

Explanation: The period T of a simple pendulum is given by T = 2π√L/g. Substituting the values, T = 2π√1.6/9.8 ≈ 2.54s.

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Correct answer: 2.54 s

237. A mass-spring system has a mass of 0.5 kg and a spring constant of 200 N/m. What is the angular frequency of the oscillation?

  • A. 10 rad/s
  • B. 14.14 rad/s
  • C. 20 rad/s
  • D. 28.28 rad/s

Explanation: The angular frequency ω is given by ω = √k/m . Substituting the values, we get ω √200/0.5 = √400 = 20rad/s.

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Correct answer: 20 rad/s

238. A coupled oscillator system consists of two identical masses m and three springs (two outer springs k, one inner spring K connecting the masses). What are the normal mode frequencies for the symmetric and antisymmetric modes?

  • A. ωsym = √k/m , ωanti = √k+2K/m
  • B. ωsym = √k+K/m , ωanti = √k/m
  • C. ωsym = √2k+K/m , ωanti = √K/m
  • D. ωsym = √k/m , ωanti = √2k + K/m

Explanation: In the symmetric mode both masses move together, so only the outer springs (k) affect the motion.

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Correct answer: ωsym = √k/m , ωanti = √k+2K/m

239. A mass of 0.5 kg is attached to a spring with spring constant 50 N/m. What is the angular frequency of oscillation?

  • A. 5 rad/s
  • B. 10 rad/s
  • C. 15 rad/s
  • D. 20 rad/s

Explanation: Angular frequency ω = √(k / m) = √(50 / 0.5) = √100 = 10 rad/s.

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Correct answer: 10 rad/s

240. A simple pendulum has a length of 1 m and oscillates with a period of 2 s. What is the acceleration due to gravity at that location?

  • A. 8.0 m/s²
  • B. 9.87 m/s²
  • C. 10.5 m/s²
  • D. 12.0 m/s²

Explanation: Period T = 2π √(L / g). Rearrange: g = 4π² L / T² = 4 × π² × 1 / 2² ≈ 39.48 / 4 ≈ 9.87 m/s².

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Correct answer: 9.87 m/s²
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