Chapter-13 Oscillations — Online MCQ Test
PHYSICS · Grade 11 · CBSE(NCERT)
Practice Chapter-13 Oscillations with a free chapter-wise online MCQ test.
This chapter covers: Periodic motion - Oscillatory motion - Period - Frequency - Simple Harmonic Motion - Phase - Force law for SHM - Energy in SHM - Simple pendulum - Free oscillations - Damped oscill....
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter-13 Oscillations — Important Questions & Answers
What is the SI unit of frequency?
- A. Hertz (Hz)
- B. Second (s)
- C. Radian per second (rad/s)
- D. Meter (m)
Answer: A. Hertz (Hz)
Frequency is the number of oscillations per unit time, and its SI unit is Hertz (Hz), equivalent to s⁻¹.
Frequency is the number of oscillations per unit time, and its SI unit is Hertz (Hz), equivalent to s⁻¹.
Which of the following is a characteristic of Simple Harmonic Motion (SHM)?
- A. Acceleration is constant
- B. Acceleration is always directed towards the mean position and is proportional to displacement
- C. Velocity remains constant throughout the motion
- D. The object moves with uniform speed
Answer: B. Acceleration is always directed towards the mean position and is proportional to displacement
In SHM, the restoring force (and hence acceleration) is directly proportional to the displacement and always directed towards the equilibrium position: a = -ω²x.
In SHM, the restoring force (and hence acceleration) is directly proportional to the displacement and always directed towards the equilibrium position: a = -ω²x.
What is the relationship between period (T) and frequency (f)?
- A. T = f
- B. T = 2πf
- C. T = 1/f
- D. T = f²
Answer: C. T = 1/f
Period is the reciprocal of frequency. If an oscillation completes f cycles per second, then each cycle takes 1/f seconds.
Period is the reciprocal of frequency. If an oscillation completes f cycles per second, then each cycle takes 1/f seconds.
The equation of SHM is given by x = A sin(ωt + φ). What does φ represent?
- A. Angular frequency
- B. Initial phase or phase constant
- C. Amplitude
- D. Time period
Answer: B. Initial phase or phase constant
The phase constant φ determines the initial state of oscillation at t = 0, indicating where in the cycle the motion begins.
The phase constant φ determines the initial state of oscillation at t = 0, indicating where in the cycle the motion begins.
In damped oscillations, what causes the gradual decrease in amplitude?
- A. Decrease in frequency
- B. Energy dissipation due to friction or resistance
- C. Increase in period
- D. Change in the equilibrium position
Answer: B. Energy dissipation due to friction or resistance
Damping occurs due to dissipative forces like friction and air resistance that remove mechanical energy from the system, reducing the amplitude over time.
Damping occurs due to dissipative forces like friction and air resistance that remove mechanical energy from the system, reducing the amplitude over time.
A mass is undergoing SHM with equation x = 5 sin(10t) cm. What is the amplitude?
- A. 10 cm
- B. 5 cm
- C. 20 cm
- D. 2.5 cm
Answer: B. 5 cm
In the equation x = A sin(ωt), the amplitude A is the coefficient of the sine function, which is 5 cm in this case.
In the equation x = A sin(ωt), the amplitude A is the coefficient of the sine function, which is 5 cm in this case.
A damped oscillator has a natural frequency ω₀ and damping coefficient b. The condition for critical damping is:
- A. b < 2mω₀
- B. b = 2mω₀
- C. b > 2mω₀
- D. b = mω₀
Answer: B. b = 2mω₀
Critical damping occurs when the damping coefficient b = 2mω₀, where the system returns to equilibrium without oscillating in minimum time.
Critical damping occurs when the damping coefficient b = 2mω₀, where the system returns to equilibrium without oscillating in minimum time.
Two identical springs are connected in series. How does the period of a mass attached to this combination compare to a single spring?
- A. Period increases by √2 times
- B. Period decreases by √2 times
- C. Period remains the same
- D. Period increases by 2 times
Answer: A. Period increases by √2 times
For springs in series, the effective spring constant k_eff = k/2. Since T = 2π√(m/k_eff), the period increases by √2.
For springs in series, the effective spring constant k_eff = k/2. Since T = 2π√(m/k_eff), the period increases by √2.
A particle executing SHM has velocity v when at displacement x. At displacement 2x, the velocity will be approximately:
- A. 2v
- B. v/2
- C. v√3
- D. √(v² - 3ω²x²)
Answer: D. √(v² - 3ω²x²)
Using energy conservation: (1/2)m(v₁²) + (1/2)kx² = (1/2)m(v₂²) + (1/2)k(2x)², solving gives v₂ = √(v² - 3ω²x²).
Using energy conservation: (1/2)m(v₁²) + (1/2)kx² = (1/2)m(v₂²) + (1/2)k(2x)², solving gives v₂ = √(v² - 3ω²x²).
In a damped SHM where the amplitude decreases as A(t) = A₀e^(-t/τ), the amplitude becomes 1/e of its initial value after time τ (called time constant). If τ = 10 s and initial amplitude is 5 cm, after how many seconds will the amplitude be 0.5 cm?
- A. 10 ln(10) s
- B. 10 ln(5) s
- C. 5 ln(10) s
- D. 20 s
Answer: A. 10 ln(10) s
Using A(t) = A₀e^(-t/τ), we get 0.5 = 5e^(-t/10), so e^(-t/10) = 0.1, giving -t/10 = ln(0.1), thus t = 10 ln(10) s.
Using A(t) = A₀e^(-t/τ), we get 0.5 = 5e^(-t/10), so e^(-t/10) = 0.1, giving -t/10 = ln(0.1), thus t = 10 ln(10) s.