Oscillations — Class 11 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 11 Physics chapter "Oscillations" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Oscillations" — 8 important questions with detailed answers for CBSE board exam prepar…
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NCERT Solutions for Class 11 Physics — Oscillations. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Oscillations — All 8 Questions Solved
Q1. A mass of 500 g is attached to a spring of spring constant k = 100 N/m. The mass is displaced 10 cm from equilibrium and released. Calculate (a) the angular frequency, (b) the period of oscillation, (c) the maximum velocity.
Step 1: Angular frequency:
ω = √(k/m)
where k = spring constant, m = mass
m = 500 g = 0.5 kg
k = 100 N/m
ω = √(100/0.5) = √200 = 14.14 rad/s
Step 2: Period of oscillation:
T = 2π/ω = 2π/14.14 = 0.444 s
Alternatively: T = 2π√(m/k) = 2π√(0.5/100) = 2π√(0.005) = 2π × 0.0707 = 0.444 s
Step 3: Maximum velocity:
v_max = ω × A
where A = amplitude = 10 cm = 0.1 m
v_max = 14.14 × 0.1 = 1.414 m/s ≈ 1.41 m/s
Final Answer: (a) Angular frequency = 14.14 rad/s; (b) Period = 0.444 s; (c) Maximum velocity = 1.41 m/s
Q2. A simple pendulum of length 1 m oscillates with amplitude 5 cm. If g = 10 m/s², calculate the maximum velocity and maximum acceleration.
Step 1: For small oscillations (amplitude << length):
Angular frequency ω = √(g/L)
L = 1 m, g = 10 m/s²
ω = √(10/1) = √10 = 3.16 rad/s
Step 2: Maximum velocity:
v_max = ω × A
where A = amplitude = 5 cm = 0.05 m
v_max = 3.16 × 0.05 = 0.158 m/s ≈ 0.16 m/s
Step 3: Maximum acceleration:
a_max = ω² × A
a_max = 10 × 0.05 = 0.5 m/s²
Alternatively:
a_max = (g/L) × A = (10/1) × 0.05 = 0.5 m/s²
Final Answer: Maximum velocity = 0.16 m/s; Maximum acceleration = 0.5 m/s²
Q3. Two identical springs are connected in series with a mass m = 200 g attached. Each spring has k = 400 N/m. Find the period of oscillation.
Step 1: For springs in series, equivalent spring constant:
1/k_eq = 1/k₁ + 1/k₂
1/k_eq = 1/400 + 1/400 = 2/400 = 1/200
k_eq = 200 N/m
Step 2: Period of oscillation:
T = 2π√(m/k_eq)
m = 200 g = 0.2 kg
k_eq = 200 N/m
T = 2π√(0.2/200)
T = 2π√(0.001)
T = 2π × 0.0316
T = 0.199 s ≈ 0.2 s
Final Answer: Period of oscillation = 0.2 s
Q4. A particle undergoes SHM with equation x = 5 sin(2πt) cm, where t is in seconds. Calculate (a) amplitude, (b) frequency, (c) period, (d) the velocity at t = 0.25 s.
Step 1: Standard SHM equation: x = A sin(ωt)
Given: x = 5 sin(2πt) cm
Amplitude A = 5 cm = 0.05 m
Step 2: Angular frequency:
ω = 2π rad/s
Step 3: Frequency:
f = ω/(2π) = 2π/(2π) = 1 Hz
Step 4: Period:
T = 1/f = 1/1 = 1 s
Step 5: Velocity in SHM:
v = dx/dt = d/dt[5 sin(2πt)] = 5 × 2π × cos(2πt) = 10π cos(2πt) cm/s
At t = 0.25 s:
v = 10π cos(2π × 0.25) = 10π cos(π/2) = 10π × 0 = 0 m/s
Final Answer: (a) Amplitude = 5 cm; (b) Frequency = 1 Hz; (c) Period = 1 s; (d) Velocity at t = 0.25 s = 0 m/s
Q5. A uniform rod of mass M and length L is pivoted at one end. Find the period of oscillation for small amplitude oscillations.
Step 1: For a physical pendulum:
T = 2π√(I/(mgd))
where I = moment of inertia about pivot, d = distance from pivot to center of mass
Step 2: For a uniform rod pivoted at one end:
I = (1/3)ML²
d = L/2 (center of mass is at midpoint)
Step 3: Substitute:
T = 2π√[(1/3)ML² / (Mg × L/2)]
T = 2π√[(1/3)ML² / (MgL/2)]
T = 2π√[(1/3)L / (g/2)]
T = 2π√[(2L)/(3g)]
T = 2π√(2L/3g)
Final Answer: Period T = 2π√(2L/3g) ≈ 2.56√(L/g)
Q6. A mass m is attached to a spring and oscillates in a viscous medium. The equation of motion is m(d²x/dt²) + b(dx/dt) + kx = 0. If b = 0.1 kg/s, k = 100 N/m, m = 1 kg, determine whether the motion is underdamped, critically damped, or overdamped.
Step 1: For damped oscillations, compare b (damping coefficient) with critical damping b_c.
Critical damping: b_c = 2√(km)
Step 2: Calculate b_c:
b_c = 2√(100 × 1) = 2√100 = 2 × 10 = 20 kg/s
Step 3: Compare given b with b_c:
Given: b = 0.1 kg/s
b_c = 20 kg/s
Since b < b_c (0.1 < 20), the motion is UNDERDAMPED
Step 4: Verify using discriminant approach:
Characteristic equation: mr² + br + k = 0
Δ = b² - 4mk = (0.1)² - 4(1)(100) = 0.01 - 400 = -399.99 < 0
Negative discriminant → complex roots → oscillatory motion with exponential decay
Final Answer: Motion is UNDERDAMPED (oscillatory with gradual decay)
Q7. Two pendulums of lengths 1 m and 2.25 m start oscillating simultaneously. After how many oscillations of the shorter pendulum will both pendulums be in phase again?
Step 1: Period of shorter pendulum (L₁ = 1 m):
T₁ = 2π√(L₁/g) = 2π√(1/10) = 2π√0.1
Step 2: Period of longer pendulum (L₂ = 2.25 m):
T₂ = 2π√(L₂/g) = 2π√(2.25/10) = 2π√0.225 = 2π × 1.5 × √0.1 = 1.5 × T₁
Step 3: To find when pendulums are in phase again:
They are in phase when the ratio of time elapsed to their periods is such that both have completed integer oscillations.
Let n₁ = number of oscillations of shorter pendulum
n₂ = number of oscillations of longer pendulum
n₁ × T₁ = n₂ × T₂
n₁ × T₁ = n₂ × 1.5T₁
n₁ = 1.5n₂
Step 4: For smallest integer values:
n₁ = 1.5 × 2 = 3 oscillations (shorter pendulum)
n₂ = 2 oscillations (longer pendulum)
Verification: 3T₁ = 2T₂ = 2 × 1.5T₁ = 3T₁ ✓
Final Answer: After 3 oscillations of shorter pendulum, both will be in phase
Q8. A spring-mass system has mass m = 0.1 kg and spring constant k = 90 N/m. The mass is given an initial velocity of 3 m/s at the equilibrium position. Calculate the amplitude of oscillation.
Step 1: At equilibrium position, all kinetic energy converts to potential energy at maximum displacement (amplitude).
Using energy conservation:
(1/2)mv₀² + (1/2)k(0)² = (1/2)m(0)² + (1/2)kA²
(1/2)mv₀² = (1/2)kA²
mv₀² = kA²
A = v₀√(m/k)
Step 2: Substitute values:
m = 0.1 kg
k = 90 N/m
v₀ = 3 m/s (initial velocity at equilibrium)
A = 3√(0.1/90)
A = 3√(1/900)
A = 3 × (1/30)
A = 3/30 = 0.1 m = 10 cm
Final Answer: Amplitude = 0.1 m or 10 cm
Oscillations — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following conditions is essential for a motion to be classified as Simple Harmonic Motion (SHM)?
- A. The acceleration is constant.
- B. The restoring force is proportional to the displacement and directed towards the equilibrium position.
- C. The velocity is always in the direction of the displacement.
- D. The potential energy is constant.
Answer: B. The restoring force is proportional to the displacement and directed towards the equilibrium position. — For SHM, the restoring force must be directly proportional to the displacement from the equilibrium position and always directed towards it (F = -kx). This leads to an acceleration proportional to displacement and opposite in direction.
MCQ 2. A particle executes SHM with an amplitude of 5 cm and a period of 4 s. What is its displacement after 1 s, if it starts from the extreme positive position?
- A. 0 cm
- B. 5 cm
- C. -5 cm
- D. 2.5 cm
Answer: A. 0 cm — If the particle starts from the extreme positive position, its displacement is given by x(t) = A cos(ωt). Here, A = 5 cm, T = 4 s, so ω = 2π/T = π/2 rad/s. At t = 1 s, x = 5 cos(π/2 * 1) = 5 cos(π/2) = 0 cm.
MCQ 3. What is the phase difference between the displacement and velocity of a particle executing Simple Harmonic Motion (SHM)?
- A. 0
- B. π/4 rad
- C. π/2 rad
- D. π rad
Answer: C. π/2 rad — If displacement is x = A sin(ωt), then velocity is v = dx/dt = Aω cos(ωt) = Aω sin(ωt + π/2). Thus, velocity leads displacement by π/2 radians.
MCQ 4. The total energy of a particle executing SHM is proportional to:
- A. Amplitude
- B. Square of the amplitude
- C. Frequency
- D. Square of the frequency
Answer: B. Square of the amplitude — The total energy (E) in SHM is given by E = (1/2)kA^2 or E = (1/2)mω^2A^2, where A is the amplitude. Therefore, total energy is proportional to the square of the amplitude.
MCQ 5. Two springs of force constants k1 and k2 are connected in series. A mass 'm' is attached to the combination. What is the equivalent force constant of the combination?
- A. k1 + k2
- B. (k1k2) / (k1 + k2)
- C. (k1 + k2) / (k1k2)
- D. sqrt(k1k2)
Answer: B. (k1k2) / (k1 + k2) — For springs connected in series, the reciprocal of the equivalent spring constant is the sum of the reciprocals of individual spring constants: 1/k_eq = 1/k1 + 1/k2. This simplifies to k_eq = (k1k2) / (k1 + k2).
MCQ 6. The time period of a simple pendulum depends on:
- A. Mass of the bob
- B. Amplitude of oscillation
- C. Length of the pendulum
- D. Material of the string
Answer: C. Length of the pendulum — For small oscillations, the time period of a simple pendulum is given by T = 2π√(L/g), which depends on the length (L) of the pendulum and acceleration due to gravity (g).
MCQ 7. A particle is executing SHM with amplitude 'A' and angular frequency 'ω'. What is its maximum velocity?
- A. Aω
- B. Aω^2
- C. A/ω
- D. A^2ω
Answer: A. Aω — The displacement in SHM is x = A sin(ωt). Differentiating with respect to time gives velocity v = Aω cos(ωt). The maximum value of cos(ωt) is 1, so the maximum velocity (v_max) is Aω.
MCQ 8. What is the maximum acceleration of a particle executing SHM with amplitude 'A' and angular frequency 'ω'?
- A. Aω
- B. Aω^2
- C. A/ω
- D. A^2ω
Answer: B. Aω^2 — The velocity in SHM is v = Aω cos(ωt). Differentiating with respect to time gives acceleration a = -Aω^2 sin(ωt). The maximum magnitude of sin(ωt) is 1, so the maximum acceleration (a_max) is Aω^2.
MCQ 9. A particle moves such that its acceleration is directly proportional to its displacement 'x' from a fixed point and is always directed towards that point. What kind of motion is it?
- A. Uniform circular motion
- B. Uniform linear motion
- C. Simple Harmonic Motion
- D. Projectile motion
Answer: C. Simple Harmonic Motion — The condition described (a = -ω^2x or F = -kx) is the defining characteristic of Simple Harmonic Motion, where the restoring force/acceleration is proportional to displacement and opposite in direction.
MCQ 10. At what position is the kinetic energy of a particle executing SHM maximum?
- A. At the extreme positions
- B. At the equilibrium position
- C. Midway between equilibrium and extreme positions
- D. Kinetic energy is constant throughout the motion.
Answer: B. At the equilibrium position — Kinetic energy is maximum when velocity is maximum. In SHM, velocity is maximum at the equilibrium position (x=0) and zero at the extreme positions.
MCQ 11. In damped oscillations, what happens to the amplitude of oscillation over time?
- A. It increases exponentially.
- B. It remains constant.
- C. It decreases exponentially.
- D. It oscillates without any specific pattern.
Answer: C. It decreases exponentially. — Damped oscillations involve energy loss due to resistive forces, causing the amplitude of oscillation to decrease exponentially over time.
MCQ 12. Resonance occurs when the driving frequency is:
- A. Much smaller than the natural frequency.
- B. Much larger than the natural frequency.
- C. Equal to the natural frequency.
- D. Independent of the natural frequency.
Answer: C. Equal to the natural frequency. — Resonance is a phenomenon where the amplitude of oscillation becomes maximum when the frequency of the external driving force matches the natural frequency of the oscillating system.
MCQ 13. A block of mass 'm' is attached to a spring with spring constant 'k'. What is the angular frequency (ω) of its oscillation?
- A. √(k/m)
- B. √(m/k)
- C. 2π√(k/m)
- D. (1/2π)√(m/k)
Answer: A. √(k/m) — For a spring-mass system, the restoring force is F = -kx. From F = ma, we get a = -(k/m)x. Comparing this with a = -ω^2x, we find ω = √(k/m).
MCQ 14. A particle is executing SHM. At what displacement from the equilibrium position is its kinetic energy equal to its potential energy?
- A. A (amplitude)
- B. A/√2
- C. A/2
- D. 0 (equilibrium position)
Answer: B. A/√2 — Total energy E = (1/2)kA^2. If KE = PE, then 2PE = E. So, 2 * (1/2)kx^2 = (1/2)kA^2, which simplifies to x^2 = A^2/2, so x = A/√2.
MCQ 15. If the frequency of an oscillating body is 10 Hz, what is its time period?
- A. 0.1 s
- B. 10 s
- C. 1 s
- D. 0.01 s
Answer: A. 0.1 s — Frequency (f) and time period (T) are reciprocals of each other, i.e., T = 1/f. So, T = 1/10 Hz = 0.1 s.
MCQ 16. How does the acceleration due to gravity 'g' affect the time period of a simple spring-mass system oscillating vertically?
- A. It increases the time period.
- B. It decreases the time period.
- C. It does not affect the time period.
- D. It makes the oscillations damped.
Answer: C. It does not affect the time period. — The time period of a spring-mass system is T = 2π√(m/k), which is independent of the acceleration due to gravity 'g'. Gravity only shifts the equilibrium position but does not change the oscillation frequency.
MCQ 17. A simple pendulum is taken from the Earth's surface to a height equal to the radius of the Earth. Its time period will become:
- A. Twice
- B. Half
- C. Four times
- D. Unchanged
Answer: A. Twice — At a height 'h' = R (Earth's radius), the effective gravity g' = g / (1 + h/R)^2 = g / (1 + 1)^2 = g/4. Since T is proportional to 1/√g, T' = T * √(g/g') = T * √(g/(g/4)) = T * √4 = 2T. The time period becomes twice.
MCQ 18. When an oscillating system vibrates with its own natural frequency without any external influence, it is called:
- A. Forced oscillation
- B. Damped oscillation
- C. Free oscillation
- D. Resonance
Answer: C. Free oscillation — Free oscillation refers to the oscillation of a system at its natural frequency when it is disturbed once and then allowed to oscillate without any external driving or damping forces.
MCQ 19. Which of the following relations is correct?
- A. ω = 2πf
- B. f = 2πT
- C. T = 2πω
- D. ω = f/2π
Answer: A. ω = 2πf — Angular frequency (ω), frequency (f), and time period (T) are related by ω = 2πf and f = 1/T. Therefore, ω = 2π/T. The correct relation among the given options is ω = 2πf.
MCQ 20. A simple pendulum is oscillating in a lift. If the lift accelerates upwards with an acceleration 'a', how does the time period of the pendulum change?
- A. It increases.
- B. It decreases.
- C. It remains the same.
- D. It becomes zero.
Answer: B. It decreases. — When the lift accelerates upwards, the effective acceleration due to gravity becomes g_eff = g + a. Since T = 2π√(L/g_eff), an increase in g_eff will cause the time period T to decrease.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.