Important Questions Class 11 Physics Chapter 13: Oscillations

Oscillatory motion is repetitive to-and-fro motion about a mean or equilibrium position.
Simple harmonic motion occurs when restoring force stays proportional to displacement and acts towards the mean position.

Oscillations explain repeated motion in pendulums, springs, musical strings, air columns, circuits and vibrating atoms. Important Questions Class 11 Physics Chapter 13 help students practise periodic motion, oscillatory motion, simple harmonic motion, phase, amplitude, angular frequency, velocity, acceleration, force law, energy in SHM and simple pendulum. The CBSE 2026 chapter also connects SHM with uniform circular motion and builds the foundation for wave motion.

Key Takeaways

• Periodic Motion: A motion that repeats after equal time intervals has time period T.
• SHM Equation: Simple harmonic displacement is x(t) = A cos(ωt + φ).
• SHM Acceleration: Acceleration follows a = −ω²x and always points towards the mean position.
• Simple Pendulum: For small angular displacement, T = 2π√(L/g).

Important Questions Class 11 Physics Chapter 13 Structure 2026

Important Questions Class 11 Physics Chapter 13 with Answers

Oscillations use periodic functions to describe repeated motion around equilibrium.
Students should identify amplitude, period, phase and restoring force before solving numericals.
These oscillations class 11 important questions follow the NCERT 2026 sequence.

1. What does Important Questions Class 11 Physics Chapter 13 mainly test?

Important Questions Class 11 Physics Chapter 13 mainly test periodic motion, SHM equations, velocity, acceleration, restoring force, energy and simple pendulum. The chapter links motion graphs with mathematical functions.

1. Periodic Motion Skill: Use T and ν correctly.
2. SHM Skill: Use x = A cos(ωt + φ).
3. Force Skill: Use F = −kx.
4. Pendulum Skill: Use T = 2π√(L/g).
5. Final Result: The chapter tests repeated motion about equilibrium.

2. Why is oscillatory motion important in physics?

Oscillatory motion is important because many physical systems move to and fro about equilibrium. Strings, pendulums, springs and atoms show oscillations.

1. Musical Strings: Vibrate to produce sound.
2. Pendulum: Oscillates about a mean position.
3. Atoms in Solids: Vibrate about equilibrium positions.
4. Final Result: Oscillations occur in mechanical, sound and atomic systems.

3. What is the main idea of Oscillations Class 11?

Oscillations Class 11 studies repeated to-and-fro motion and its mathematical description. It focuses on time period, frequency, amplitude, phase and SHM.

1. Repeated Motion: Motion repeats after a fixed time.
2. Equilibrium: Motion occurs about a mean position.
3. Mathematics: Sine and cosine functions describe SHM.
4. Final Result: Oscillations describe periodic motion around equilibrium.

Periodic Motion Class 11 Questions

Periodic motion repeats itself after a fixed time interval.
It may or may not involve motion about a mean position.
These periodic motion class 11 questions explain period, frequency and examples.

4. What is periodic motion?

Periodic motion is motion that repeats itself after equal intervals of time. The smallest repeat time is called the period.

1. Repeat Condition: Motion repeats identically.
2. Time Interval: Fixed for each cycle.
3. Example: Earth’s rotation about its axis.
4. Final Result: Periodic motion repeats after a fixed period.

5. What is time period?

Time period is the smallest time after which a periodic motion repeats itself. Its SI unit is second.

1. Symbol: T.
2. Meaning: Time for one complete cycle.
3. Unit: Second.
4. Final Result: Time period measures one full repetition.

6. What is frequency of periodic motion?

Frequency is the number of repetitions per unit time. Its SI unit is hertz.

1. Symbol: ν.
2. Formula Used: ν = 1/T.
3. Unit: Hz or s^-1.
4. Final Result: Frequency equals reciprocal of time period.

7. What is the relation between frequency and time period?

The relation is ν = 1/T. A shorter period means higher frequency.

1. Time Period: T.
2. Frequency: ν.
3. Formula: ν = 1/T.
4. Final Result: Frequency and period are reciprocals.

8. A heart beats 75 times per minute. Find frequency and period.

The frequency is 1.25 Hz, and the period is 0.8 s. Convert one minute into 60 seconds.

1. Given Data:
   Beats = 75
   Time = 60 s
2. Frequency:
   ν = 75/60
   ν = 1.25 Hz
3. Period:
   T = 1/ν = 1/1.25
   T = 0.8 s
4. Final Result: Frequency = 1.25 Hz and period = 0.8 s.

9. Is every periodic motion oscillatory?

No, every periodic motion is not oscillatory. Uniform circular motion is periodic but not to-and-fro motion about a mean position.

1. Periodic Motion: Repeats after a fixed time.
2. Oscillatory Motion: Moves about equilibrium.
3. Example: Circular motion repeats without to-and-fro motion.
4. Final Result: Periodic motion need not be oscillatory.

Oscillatory Motion Class 11 Questions

Oscillatory motion is a special kind of periodic motion.
The body moves to and fro about a stable equilibrium position.
These oscillatory motion class 11 questions cover equilibrium, restoring force and vibration.

10. What is oscillatory motion?

Oscillatory motion is repeated to-and-fro motion about a mean position. A restoring force brings the body back towards equilibrium.

1. Mean Position: Equilibrium position.
2. Motion Type: To-and-fro.
3. Force: Directed towards mean position.
4. Final Result: Oscillatory motion occurs about equilibrium.

11. What is equilibrium position in oscillations?

Equilibrium position is the position where net external force on the body is zero. If left at rest there, the body remains there.

1. Net Force: Zero.
2. Rest Condition: Body remains at rest.
3. Example: Bottom point of a ball in a bowl.
4. Final Result: Equilibrium position has zero net force.

12. What is the difference between oscillation and vibration?

Oscillation usually refers to low-frequency to-and-fro motion, while vibration often refers to high-frequency oscillation. The physical idea is same.

1. Oscillation Example: Branch of a tree.
2. Vibration Example: Guitar string.
3. Frequency Difference: Usage changes with frequency.
4. Final Result: Both describe repeated motion about equilibrium.

13. Why do real oscillations eventually stop?

Real oscillations stop because damping removes mechanical energy. Friction and air resistance convert energy into heat.

1. Damping Cause: Friction and resistance.
2. Energy Loss: Mechanical energy decreases.
3. Final State: Body rests at equilibrium.
4. Final Result: Damping makes real oscillations die out.

14. What is displacement in oscillatory motion?

Displacement is the change of the oscillating variable from its reference or equilibrium value. It can be positive or negative.

1. Spring Block: Displacement from equilibrium position.
2. Pendulum: Angular displacement from vertical.
3. Sound Wave: Pressure variation from mean pressure.
4. Final Result: Displacement is measured from equilibrium.

Simple Harmonic Motion Class 11 Questions

Simple harmonic motion is the simplest oscillatory motion.
Its displacement varies sinusoidally with time.
These simple harmonic motion class 11 questions cover equation, amplitude, phase and angular frequency.

15. What is simple harmonic motion?

Simple harmonic motion is oscillatory motion in which displacement is a sine or cosine function of time. It is written as x = A cos(ωt + φ).

1. Displacement: x.
2. Amplitude: A.
3. Phase: ωt + φ.
4. Final Result: SHM has sinusoidal displacement.

16. What is the displacement equation of SHM?

The displacement equation of SHM is x(t) = A cos(ωt + φ). It gives position as a function of time.

1. A: Amplitude.
2. ω: Angular frequency.
3. φ: Phase constant.
4. Final Result: SHM displacement is x(t) = A cos(ωt + φ).

17. What is amplitude in SHM?

Amplitude is the maximum displacement from mean position. It is denoted by A.

1. Maximum Positive Displacement: +A.
2. Maximum Negative Displacement: −A.
3. Total Separation of Extremes: 2A.
4. Final Result: Amplitude is maximum displacement from equilibrium.

18. What is phase in SHM?

Phase is the quantity that decides the state of motion at a given time. For SHM, phase is ωt + φ.

1. Time-dependent Part: ωt.
2. Initial Part: φ.
3. Use: Determines displacement and velocity at time t.
4. Final Result: Phase specifies the stage of SHM.

19. What is phase constant?

Phase constant is the phase at t = 0. It depends on the initial position and velocity.

1. Symbol: φ.
2. At t = 0: Phase = φ.
3. Role: Sets starting point of oscillation.
4. Final Result: Phase constant fixes initial condition.

20. What is angular frequency in SHM?

Angular frequency is the rate of change of phase. It is related to time period by ω = 2π/T.

1. Symbol: ω.
2. Unit: rad s^-1.
3. Relation: ω = 2πν.
4. Final Result: Angular frequency equals 2π times frequency.

SHM Class 11 Important Questions

SHM appears when the restoring force is proportional to displacement and directed towards the mean position.
The same motion can be described using displacement equation or force law.
These SHM class 11 important questions focus on standard exam checks.

21. What conditions are required for SHM?

SHM requires a restoring force proportional to displacement and directed towards equilibrium. The force law is F = −kx.

1. Proportionality: F ∝ x.
2. Direction: Opposite to displacement.
3. Mean Position: Force points towards equilibrium.
4. Final Result: SHM follows restoring force F = −kx.

22. Is every oscillatory motion simple harmonic?

No, every oscillatory motion is not simple harmonic. SHM needs displacement to follow a sine or cosine function.

1. Oscillatory Motion: Repeats about equilibrium.
2. SHM Condition: Restoring force must be linear.
3. Example: Large-angle pendulum is not exact SHM.
4. Final Result: Only linear restoring oscillations are SHM.

23. Which equation represents SHM, a = −10x or a = 0.7x?

The equation a = −10x represents SHM. Acceleration must be proportional and opposite to displacement.

1. SHM Form: a = −ω²x.
2. Equation a = −10x: Matches SHM form.
3. Equation a = 0.7x: Acceleration points away from mean.
4. Final Result: a = −10x represents SHM.

24. Does a = −200x² represent SHM?

No, a = −200x² does not represent SHM. Acceleration is not linearly proportional to displacement.

1. SHM Requirement: a ∝ −x.
2. Given Relation: a ∝ −x².
3. Linearity: Not satisfied.
4. Final Result: a = −200x² is not SHM.

25. If x = 5 cos(2πt + π/4), find amplitude and period.

The amplitude is 5 m, and period is 1 s. Compare with x = A cos(ωt + φ).

1. Given Equation: x = 5 cos(2πt + π/4).
2. Amplitude: A = 5 m.
3. Angular Frequency: ω = 2π s^-1.
4. Period: T = 2π/ω = 1 s.
5. Final Result: A = 5 m and T = 1 s.

26. For x = 5 cos(2πt + π/4), find displacement at t = 1.5 s.

The displacement is −3.535 m. Substitute t = 1.5 s.

1. Formula: x = 5 cos(2πt + π/4).
2. Substitution:
   x = 5 cos(3π + π/4)
3. Value:
   cos(13π/4) = −1/√2
4. Final Result: x = −3.535 m.

Uniform Circular Motion and SHM Class 11 Questions

SHM can be understood as the projection of uniform circular motion on a diameter.
This connection helps explain sinusoidal displacement, velocity and acceleration.
These uniform circular motion and SHM class 11 questions cover reference circle logic.

27. How is SHM related to uniform circular motion?

SHM is the projection of uniform circular motion on a diameter. The projected point moves to and fro sinusoidally.

1. Reference Motion: Uniform circular motion.
2. Projection: Taken on a diameter.
3. Result: x = A cos(ωt + φ).
4. Final Result: Projection of circular motion gives SHM.

28. What is a reference circle in SHM?

A reference circle is the circle used to represent SHM through uniform circular motion. Its radius equals the amplitude.

1. Radius: A.
2. Angular Speed: ω.
3. Projection: Gives SHM displacement.
4. Final Result: Reference circle links circular motion and SHM.

29. Why does projection of circular motion become SHM?

Projection becomes SHM because the coordinate of a uniformly rotating radius varies as sine or cosine. This matches SHM displacement.

1. Rotating Radius: Makes angle ωt + φ.
2. Projection: x = A cos(ωt + φ).
3. SHM Form: Same displacement equation.
4. Final Result: Sinusoidal projection creates SHM.

30. If radius A rotates anticlockwise with period 4 s and initial phase π/4, write x(t).

The SHM is x(t) = A cos(πt/2 + π/4). Use ω = 2π/T.

1. Given Data:
   T = 4 s
   φ = π/4
2. Angular Frequency:
   ω = 2π/4 = π/2 rad s^-1
3. Displacement:
   x(t) = A cos(ωt + φ)
4. Final Result: x(t) = A cos(πt/2 + π/4).

31. If circular motion has period 30 s and starts from y-axis clockwise, write x(t).

The projected SHM is x(t) = B sin(πt/15). It has amplitude B and period 30 s.

1. Period: T = 30 s.
2. Angular Frequency: ω = 2π/30 = π/15.
3. Initial x-coordinate: 0.
4. Clockwise Projection: x(t) = B sin(πt/15).
5. Final Result: x(t) = B sin(πt/15).

Velocity and Acceleration in SHM Class 11

Velocity and acceleration in SHM also vary sinusoidally.
Velocity is maximum at mean position, while acceleration is maximum at extreme positions.
These velocity and acceleration in SHM class 11 questions cover signs, phases and formulas.

32. What is velocity in SHM?

Velocity in SHM is v(t) = −ωA sin(ωt + φ) when x = A cos(ωt + φ). It is the time derivative of displacement.

1. Displacement: x = A cos(ωt + φ).
2. Differentiate: v = dx/dt.
3. Formula: v = −ωA sin(ωt + φ).
4. Final Result: Velocity is sinusoidal in SHM.

33. What is acceleration in SHM?

Acceleration in SHM is a(t) = −ω²x(t). It always points towards the mean position.

1. Velocity: v = −ωA sin(ωt + φ).
2. Differentiate: a = dv/dt.
3. Formula: a = −ω²A cos(ωt + φ).
4. Final Result: Acceleration is proportional to −x.

34. Where is velocity maximum in SHM?

Velocity is maximum at the mean position. Its maximum value is vmax = ωA.

1. Mean Position: x = 0.
2. Energy Form: All energy is kinetic.
3. Velocity: Maximum magnitude.
4. Final Result: Maximum speed equals ωA.

35. Where is velocity zero in SHM?

Velocity is zero at the extreme positions. These positions are x = +A and x = −A.

1. Extreme Displacement: Maximum magnitude.
2. Turning Point: Direction changes.
3. Speed: Zero at that instant.
4. Final Result: Velocity is zero at extremes.

36. Where is acceleration maximum in SHM?

Acceleration is maximum at the extreme positions. Its magnitude is amax = ω²A.

1. Formula: a = −ω²x.
2. At Extremes: |x| = A.
3. Magnitude: |a| = ω²A.
4. Final Result: Maximum acceleration equals ω²A.

37. Where is acceleration zero in SHM?

Acceleration is zero at the mean position. At this point, displacement is zero.

1. Mean Position: x = 0.
2. Formula: a = −ω²x.
3. Result: a = 0.
4. Final Result: Acceleration is zero at equilibrium.

38. For x = 5 cos(2πt + π/4), find speed at t = 1.5 s.

The speed is 22 m/s approximately. Use v = −ωA sin(ωt + φ).

1. Given Data:
   A = 5 m
   ω = 2π s^-1
   t = 1.5 s
2. Formula:
   v = −10π sin(3π + π/4)
3. Calculation:
   |v| = 10π/√2
   |v| ≈ 22 m/s
4. Final Result: Speed ≈ 22 m/s.

39. For the same SHM, find acceleration at t = 1.5 s.

The acceleration is 140 m/s² approximately. Use a = −ω²x.

1. Given Data:
   x = −3.535 m
   ω = 2π s^-1
2. Formula Used: a = −ω²x.
3. Calculation:
   a = −(2π)²(−3.535)
   a ≈ 140 m/s²
4. Final Result: Acceleration ≈ 140 m/s².

Force Law for SHM Class 11 Questions

The force law of SHM gives the physical condition for simple harmonic motion.
It states that restoring force acts opposite to displacement.
These force law for SHM class 11 questions cover springs and linear oscillators.

40. What is the force law for SHM?

The force law for SHM is F = −kx. The negative sign shows that force acts towards the mean position.

1. Restoring Force: F.
2. Displacement: x.
3. Force Constant: k.
4. Final Result: SHM needs F = −kx.

41. How is angular frequency related to spring constant?

Angular frequency is ω = √(k/m) for a spring-mass oscillator. Here m is the mass.

1. Force Law: F = −kx.
2. Newton’s Law: F = ma.
3. SHM Form: a = −(k/m)x.
4. Final Result: ω = √(k/m).

42. What is the time period of a spring-mass system?

The time period is T = 2π√(m/k). It does not depend on amplitude for ideal SHM.

1. Mass: m.
2. Spring Constant: k.
3. Formula Used: T = 2π√(m/k).
4. Final Result: Spring time period depends on m and k.

43. Two identical springs of constant k pull a mass m from both sides. Find period.

The period is T = 2π√(m/2k). The effective restoring constant is 2k.

1. Left Spring Force: −kx.
2. Right Spring Force: −kx.
3. Net Force: F = −2kx.
4. Effective Constant: keff = 2k.
5. Final Result: T = 2π√(m/2k).

44. A spring has k = 1200 N/m and mass 3 kg. Find angular frequency.

The angular frequency is 20 rad/s. Use ω = √(k/m).

1. Given Data:
   k = 1200 N/m
   m = 3 kg
2. Formula Used: ω = √(k/m).
3. Calculation:
   ω = √(1200/3)
   ω = √400
   ω = 20 rad/s
4. Final Result: ω = 20 rad/s.

45. For the same spring, find frequency.

The frequency is 3.18 Hz approximately. Use ν = ω/2π.

1. Given Data:
   ω = 20 rad/s
2. Formula Used: ν = ω/2π.
3. Calculation:
   ν = 20/(2π)
   ν ≈ 3.18 Hz
4. Final Result: Frequency ≈ 3.18 Hz.

Energy in SHM Class 11 Questions

Energy in SHM changes form between kinetic and potential energy.
Total mechanical energy remains constant when damping is absent.
These energy in SHM class 11 questions cover K, U and total energy.

46. What is kinetic energy in SHM?

Kinetic energy in SHM is K = 1/2 mv². It is maximum at mean position and zero at extremes.

1. Velocity: v.
2. Mass: m.
3. Formula Used: K = 1/2 mv².
4. Final Result: Kinetic energy depends on speed squared.

47. What is potential energy in SHM?

Potential energy in SHM is U = 1/2 kx². It is maximum at extreme positions.

1. Force Constant: k.
2. Displacement: x.
3. Formula Used: U = 1/2 kx².
4. Final Result: Potential energy depends on x².

48. What is total energy in SHM?

Total energy in SHM is E = 1/2 kA². It remains constant without damping.

1. Amplitude: A.
2. Force Constant: k.
3. Formula Used: E = 1/2 kA².
4. Final Result: Total energy is proportional to amplitude squared.

49. Where is total energy located at mean position?

At mean position, total energy is entirely kinetic. Potential energy is zero because x = 0.

1. Mean Position: x = 0.
2. Potential Energy: U = 0.
3. Kinetic Energy: Maximum.
4. Final Result: At mean position, energy is kinetic.

50. Where is total energy located at extreme positions?

At extreme positions, total energy is entirely potential. Kinetic energy is zero because velocity is zero.

1. Extreme Position: x = ±A.
2. Velocity: v = 0.
3. Potential Energy: Maximum.
4. Final Result: At extremes, energy is potential.

51. A 1 kg block on spring k = 50 N/m has amplitude 10 cm. Find total energy.

The total energy is 0.25 J. Use E = 1/2 kA².

1. Given Data:
   m = 1 kg
   k = 50 N/m
   A = 10 cm = 0.10 m
2. Formula Used: E = 1/2 kA².
3. Calculation:
   E = 1/2 × 50 × 0.10²
   E = 0.25 J
4. Final Result: Total energy = 0.25 J.

52. For the same block, find potential energy at x = 5 cm.

The potential energy is 0.0625 J. Use U = 1/2 kx².

1. Given Data:
   k = 50 N/m
   x = 5 cm = 0.05 m
2. Formula Used: U = 1/2 kx².
3. Calculation:
   U = 1/2 × 50 × 0.05²
   U = 0.0625 J
4. Final Result: Potential energy = 0.0625 J.

53. For the same block, find kinetic energy at x = 5 cm.

The kinetic energy is 0.1875 J. Use K = E − U.

1. Total Energy: E = 0.25 J.
2. Potential Energy: U = 0.0625 J.
3. Formula Used: K = E − U.
4. Calculation:
   K = 0.25 − 0.0625
   K = 0.1875 J
5. Final Result: Kinetic energy = 0.1875 J.

Simple Pendulum Class 11 Questions

A simple pendulum performs approximate SHM for small angular displacement.
Its time period depends on length and acceleration due to gravity.
These simple pendulum class 11 questions cover restoring torque and small-angle approximation.

54. What is a simple pendulum?

A simple pendulum consists of a small bob suspended by a light inextensible string from a fixed support. It oscillates under gravity.

1. Bob: Small heavy mass.
2. String: Light and inextensible.
3. Motion: To-and-fro about vertical mean position.
4. Final Result: A simple pendulum oscillates under gravity.

55. When does a simple pendulum perform SHM?

A simple pendulum performs SHM for small angular displacement. In this case, sin θ ≈ θ in radians.

1. Restoring Torque: τ = −mgL sin θ.
2. Small Angle: sin θ ≈ θ.
3. SHM Form: Angular acceleration ∝ −θ.
4. Final Result: Small-angle pendulum motion is SHM.

56. What is the time period of a simple pendulum?

The time period of a simple pendulum is T = 2π√(L/g). It applies for small oscillations.

1. Length: L.
2. Acceleration Due to Gravity: g.
3. Formula Used: T = 2π√(L/g).
4. Final Result: Pendulum period depends on L and g.

57. Does simple pendulum period depend on mass?

No, simple pendulum period does not depend on bob mass. The formula T = 2π√(L/g) has no mass term.

1. Formula: T = 2π√(L/g).
2. Mass Term: Absent.
3. Condition: Small oscillations.
4. Final Result: Pendulum period is independent of mass.

58. Does simple pendulum period depend on amplitude?

For small oscillations, simple pendulum period does not depend on amplitude. Large angular displacement breaks the approximation.

1. Small Angle: sin θ ≈ θ.
2. Formula: T = 2π√(L/g).
3. Amplitude: Not present in formula.
4. Final Result: Small-amplitude pendulum period is amplitude independent.

Time Period of Simple Pendulum Class 11

The time period formula helps compare pendulums on Earth, Moon and moving systems.
A seconds pendulum has time period 2 seconds.
These time period of simple pendulum class 11 questions support direct numerical practice.

59. What is a seconds pendulum?

A seconds pendulum is a pendulum with time period 2 s. It ticks once every second from one extreme to the other.

1. One Complete Oscillation: 2 s.
2. Half Oscillation: 1 s.
3. Name: Seconds pendulum.
4. Final Result: Seconds pendulum has T = 2 s.

60. Find length of a seconds pendulum on Earth.

The length is about 1 m. Use T = 2π√(L/g).

1. Given Data:
   T = 2 s
   g = 9.8 m/s²
2. Formula Used: L = gT²/(4π²).
3. Calculation:
   L = 9.8 × 4/(4π²)
   L ≈ 1 m
4. Final Result: Length of seconds pendulum ≈ 1 m.

61. A pendulum has T = 3.5 s on Earth. Find its period on Moon where g = 1.7 m/s².

The period on Moon is 8.4 s approximately. Use T ∝ 1/√g for same length.

1. Given Data:
   TE = 3.5 s
   gE = 9.8 m/s²
   gM = 1.7 m/s²
2. Formula Used: TM/TE = √(gE/gM).
3. Calculation:
   TM = 3.5√(9.8/1.7)
   TM ≈ 8.4 s
4. Final Result: Moon period ≈ 8.4 s.

62. What happens to pendulum time period if length becomes four times?

The time period becomes two times. It is proportional to square root of length.

1. Formula: T = 2π√(L/g).
2. Length Change: L becomes 4L.
3. Period Change: T becomes √4T.
4. Final Result: Period doubles when length becomes four times.

63. What happens to pendulum time period if g decreases?

The time period increases when g decreases. It is inversely proportional to √g.

1. Formula: T = 2π√(L/g).
2. Relation: T ∝ 1/√g.
3. Example: Pendulum runs slower on Moon.
4. Final Result: Lower g gives longer period.

NCERT Class 11 Physics Chapter 13 Questions

NCERT questions test classification, SHM equations, spring systems, circular projection and pendulum formulas.
Students should check whether motion is periodic, oscillatory or simple harmonic.
These NCERT Class 11 Physics Chapter 13 questions follow the 2026 exercise pattern.

64. Which examples represent periodic motion?

Periodic motion includes a swimmer’s return trip, a freely suspended bar magnet and rotating hydrogen molecule. An arrow released from a bow is not periodic.

1. Swimmer Return Trip: Repeats when motion is repeated.
2. Bar Magnet: Oscillates about magnetic meridian.
3. Hydrogen Molecule Rotation: Periodic rotation.
4. Final Result: Arrow motion is non-periodic.

65. Which examples are nearly simple harmonic?

Oscillating mercury in a U-tube and a ball in a smooth bowl near the bottom are nearly SHM. Earth’s rotation is periodic but not SHM.

1. U-tube Mercury: Restoring force approximately proportional to displacement.
2. Smooth Bowl: Small displacement gives restoring force.
3. Earth Rotation: Periodic but not to-and-fro.
4. Final Result: SHM needs restoring force proportional to displacement.

66. A particle moves between A and B, 10 cm apart. What is amplitude?

The amplitude is 5 cm. The mean position lies midway between A and B.

1. Distance AB: 10 cm.
2. Mean Position: Midpoint of AB.
3. Amplitude: Half of total separation.
4. Final Result: Amplitude = 5 cm.

67. In the same motion, what is velocity at the extreme point A?

The velocity at extreme point A is zero. The particle reverses direction there.

1. Extreme Point: Turning point.
2. Speed: Momentarily zero.
3. Force: Directed towards mean position.
4. Final Result: Velocity at an extreme is zero.

68. In the same motion, what is acceleration at the midpoint?

The acceleration at midpoint is zero. The midpoint is the equilibrium position.

1. Midpoint: Mean position.
2. Displacement: x = 0.
3. Formula: a = −ω²x.
4. Final Result: Acceleration at mean position is zero.

69. A spring balance reads up to 50 kg over 20 cm. A body oscillates with period 0.6 s. Find its mass.

The body’s mass is about 22.3 kg. Use scale extension to find spring constant.

1. Maximum Load: 50 kg gives force = 50g = 490 N.
2. Scale Length: x = 0.20 m.
3. Spring Constant: k = 490/0.20 = 2450 N/m.
4. Period Formula: T = 2π√(m/k).
5. Calculation:
   m = kT²/(4π²)
   m = 2450 × 0.36/(4π²)
   m ≈ 22.3 kg
6. Final Result: Mass ≈ 22.3 kg.

70. A spring has k = 1200 N/m, mass 3 kg and amplitude 2 cm. Find maximum acceleration.

The maximum acceleration is 8 m/s². Use amax = ω²A.

1. Given Data:
   k = 1200 N/m
   m = 3 kg
   A = 2 cm = 0.02 m
2. Angular Frequency:
   ω² = k/m = 1200/3 = 400
3. Formula Used: amax = ω²A.
4. Calculation:
   amax = 400 × 0.02
   amax = 8 m/s²
5. Final Result: Maximum acceleration = 8 m/s².

71. For the same spring, find maximum speed.

The maximum speed is 0.4 m/s. Use vmax = ωA.

1. Angular Frequency: ω = 20 rad/s.
2. Amplitude: A = 0.02 m.
3. Formula Used: vmax = ωA.
4. Calculation:
   vmax = 20 × 0.02
   vmax = 0.4 m/s
5. Final Result: Maximum speed = 0.4 m/s.

Class 11 Physics Chapter 13 Numericals

Numericals in this chapter often combine angular frequency, spring constant, amplitude and time period.
Students should keep angle values in radians unless degrees are clearly stated.
These Class 11 Physics Chapter 13 numericals support CBSE 2026 practice.

72. Find period if angular frequency is 5 rad/s.

The period is 2π/5 s or about 1.26 s. Use T = 2π/ω.

1. Given Data:
   ω = 5 rad/s
2. Formula Used: T = 2π/ω.
3. Calculation:
   T = 2π/5
   T ≈ 1.26 s
4. Final Result: T ≈ 1.26 s.

73. Find angular frequency if frequency is 4 Hz.

The angular frequency is 8π rad/s. Use ω = 2πν.

1. Given Data:
   ν = 4 Hz
2. Formula Used: ω = 2πν.
3. Calculation:
   ω = 2π × 4
   ω = 8π rad/s
4. Final Result: ω = 8π rad/s.

74. A particle has amplitude 0.2 m and angular frequency 10 rad/s. Find maximum speed.

The maximum speed is 2 m/s. Use vmax = ωA.

1. Given Data:
   A = 0.2 m
   ω = 10 rad/s
2. Formula Used: vmax = ωA.
3. Calculation:
   vmax = 10 × 0.2
   vmax = 2 m/s
4. Final Result: Maximum speed = 2 m/s.

75. For the same particle, find maximum acceleration.

The maximum acceleration is 20 m/s². Use amax = ω²A.

1. Given Data:
   A = 0.2 m
   ω = 10 rad/s
2. Formula Used: amax = ω²A.
3. Calculation:
   amax = 10² × 0.2
   amax = 20 m/s²
4. Final Result: Maximum acceleration = 20 m/s².

76. A mass of 0.5 kg oscillates on a spring of k = 200 N/m. Find time period.

The time period is 0.314 s approximately. Use T = 2π√(m/k).

1. Given Data:
   m = 0.5 kg
   k = 200 N/m
2. Formula Used: T = 2π√(m/k).
3. Calculation:
   T = 2π√(0.5/200)
   T = 2π√0.0025
   T = 0.314 s
4. Final Result: T ≈ 0.314 s.

77. A pendulum length is 0.25 m. Find period using g = 9.8 m/s².

The period is about 1.00 s. Use T = 2π√(L/g).

1. Given Data:
   L = 0.25 m
   g = 9.8 m/s²
2. Formula Used: T = 2π√(L/g).
3. Calculation:
   T = 2π√(0.25/9.8)
   T ≈ 1.00 s
4. Final Result: Period ≈ 1.00 s.

CBSE Class 11 Physics Chapter-Wise Important Questions