ECAT (Engineering College Admission) Physics Oscillations — Set 2

Oscillations MCQs set 2 for ECAT (Engineering College Admission) Physics — 20 solved questions.

ECAT (Engineering College Admission) Physics Oscillations — Set 2

  1. Question 1

    Q1. The motion of a simple pendulum is an example of

    • A) Simple harmonic motion
    • B) Circular motion
    • C) Rotatory motion
    • D) Translatory motion

    Answer: Simple harmonic motion

    Explanation: Pendulum's motion is periodic and follows SHM equation d²θ/dt² = - (g/l) θ, where g is acceleration due to gravity and l is length.

  2. Question 2

    Q2. The time period of a simple pendulum is given by

    • A) 2π √(l/g)
    • B) 2π √(g/l)
    • C) 2π √(l/g) ³
    • D) 1 / (2π √(l/g))

    Answer: 2π √(l/g)

    Explanation: Derived from SHM equation, time period T = 2π √(l/g), where l is pendulum length and g is acceleration due to gravity.

  3. Question 3

    Q3. The frequency of oscillation of a mass-spring system is

    • A) (1/2π) √(k/m)
    • B) (1/2π) √(m/k)
    • C) 2π √(k/m)
    • D) 2π √(m/k)

    Answer: (1/2π) √(k/m)

    Explanation: Frequency f = (1/2π) √(k/m), where k is spring constant and m is mass attached to the spring.

  4. Question 4

    Q4. The total energy of a simple harmonic oscillator is

    • A) Proportional to amplitude
    • B) Proportional to amplitude ²
    • C) Proportional to frequency
    • D) Proportional to frequency ²

    Answer: Proportional to amplitude ²

    Explanation: Total energy E = (1/2) m ω² A², where ω is angular frequency and A is amplitude, so E ∝ A².

  5. Question 5

    Q5. The equation of simple harmonic motion is given by

    • A) d²x/dt² + ω² x = 0
    • B) d²x/dt² - ω² x = 0
    • C) dx/dt + ω x = 0
    • D) dx/dt - ω x = 0

    Answer: d²x/dt² + ω² x = 0

    Explanation: SHM is represented by d²x/dt² + ω² x = 0, where ω is angular frequency and x is displacement.

  6. Question 6

    Q6. The angular frequency of a mass-spring system is

    • A) √(k/m)
    • B) √(m/k)
    • C) k/m
    • D) m/k

    Answer: √(k/m)

    Explanation: Angular frequency ω = √(k/m), where k is spring constant and m is mass.

  7. Question 7

    Q7. The time period of a physical pendulum is given by

    • A) 2π √(I/mgd)
    • B) 2π √(m/Igd)
    • C) 2π √(I/mg)
    • D) 2π √(mg/I)

    Answer: 2π √(I/mgd)

    Explanation: Time period T = 2π √(I/mgd), where I is moment of inertia, m is mass, g is acceleration due to gravity, and d is distance from pivot to center of mass.

  8. Question 8

    Q8. The amplitude of a damped oscillator decreases exponentially with

    • A) Time
    • B) Frequency
    • C) Displacement
    • D) Velocity

    Answer: Time

    Explanation: Amplitude A(t) = A₀ e^(-bt), where b is damping coefficient, so amplitude decreases exponentially with time.

  9. Question 9

    Q9. The resonance occurs when the frequency of the external force is

    • A) Equal to the natural frequency
    • B) Greater than the natural frequency
    • C) Less than the natural frequency
    • D) Zero

    Answer: Equal to the natural frequency

    Explanation: Resonance occurs when driving frequency matches natural frequency, maximizing amplitude.

  10. Question 10

    Q10. The quality factor of a damped oscillator is given by

    • A) ω₀ / (2b)
    • B) ω₀ / b
    • C) b / ω₀
    • D) 2b / ω₀

    Answer: ω₀ / (2b)

    Explanation: Quality factor Q = ω₀ / (2b), where ω₀ is natural frequency and b is damping coefficient.

  11. Question 11

    Q11. The equation of a damped oscillator is given by

    • A) d²x/dt² + 2b dx/dt + ω₀² x = 0
    • B) d²x/dt² - 2b dx/dt + ω₀² x = 0
    • C) d²x/dt² + 2b dx/dt - ω₀² x = 0
    • D) d²x/dt² - 2b dx/dt - ω₀² x = 0

    Answer: d²x/dt² + 2b dx/dt + ω₀² x = 0

    Explanation: Damped oscillator is represented by d²x/dt² + 2b dx/dt + ω₀² x = 0, where b is damping coefficient and ω₀ is natural frequency.

  12. Question 12

    Q12. The energy of a simple harmonic oscillator is proportional to

    • A) Amplitude
    • B) Amplitude ²
    • C) Frequency
    • D) Frequency ²

    Answer: Amplitude ²

    Explanation: Total energy E = (1/2) m ω² A², so E ∝ A², where A is amplitude.

  13. Question 13

    Q13. The maximum velocity of a simple harmonic oscillator is

    • A) Aω
    • B) Aω²
    • C) A/ω
    • D) A/ω²

    Answer: Aω

    Explanation: Maximum velocity v_max = Aω, where A is amplitude and ω is angular frequency.

  14. Question 14

    Q14. The maximum acceleration of a simple harmonic oscillator is

    • A) Aω²
    • B) Aω
    • C) A/ω
    • D) A/ω²

    Answer: Aω²

    Explanation: Maximum acceleration a_max = Aω², where A is amplitude and ω is angular frequency.

  15. Question 15

    Q15. The simple harmonic motion is represented by

    • A) x = A sin(ωt + φ)
    • B) x = A cos(ωt + φ)
    • C) x = A sin(ωt)
    • D) Both A and B

    Answer: Both A and B

    Explanation: SHM can be represented by x = A sin(ωt + φ) or x = A cos(ωt + φ), both are valid representations.

  16. Question 16

    Q16. The frequency of a simple pendulum on the surface of the earth and at height h above it are related as

    • A) f_h = f₀ √(R / (R + h))
    • B) f_h = f₀ √(R + h) / R
    • C) f_h = f₀ (R / (R + h))
    • D) f_h = f₀ (R + h) / R

    Answer: f_h = f₀ √(R / (R + h))

    Explanation: Frequency f ∝ √g, and g at height h is g_h = g₀ (R / (R + h))², so f_h = f₀ √(R / (R + h)).

  17. Question 17

    Q17. The time period of a simple pendulum in a lift accelerating upwards is

    • A) 2π √(l / (g + a))
    • B) 2π √(l / (g - a))
    • C) 2π √(l / g)
    • D) 2π √(l / a)

    Answer: 2π √(l / (g + a))

    Explanation: Effective g is g + a when lift accelerates upwards, so T = 2π √(l / (g + a)).

  18. Question 18

    Q18. The time period of a simple pendulum in a lift accelerating downwards is

    • A) 2π √(l / (g - a))
    • B) 2π √(l / (g + a))
    • C) 2π √(l / g)
    • D) 2π √(l / a)

    Answer: 2π √(l / (g - a))

    Explanation: Effective g is g - a when lift accelerates downwards, so T = 2π √(l / (g - a)).

  19. Question 19

    Q19. The graph between time period and length of a simple pendulum is

    • A) Parabolic
    • B) Linear
    • C) Hyperbolic
    • D) None of these

    Answer: Parabolic

    Explanation: T = 2π √(l/g), so T² ∝ l, a parabolic relation between T and √l.

  20. Question 20

    Q20. A simple pendulum has a time period T. What will be its new time period if its length is doubled?

    • A) T √2
    • B) T / √2
    • C) √(2T)
    • D) T √(1/2)

    Answer: T √2

    Explanation: T = 2π √(l/g). When l is doubled, T becomes √2 times. So, new T = T √2.