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Oscillations

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Chapter Notes

Overview
This chapter introduces oscillatory motion, a special type of periodic motion where an object moves to and fro about a mean position. The central idea is Simple Harmonic Motion (SHM), the simplest form of oscillation, where the restoring force is directly proportional to the displacement and acts towards the mean position. A student must understand that SHM can be described by sinusoidal functions, linked to uniform circular motion, and characterized by its period, frequency, amplitude, phase, and energy conservation.
Key Concepts (13)
Periodic motion
A motion that repeats itself at regular intervals of time.
e.g. The orbital motion of planets in the solar system.
Oscillatory motion
A periodic to and fro motion about a mean position (equilibrium position).
e.g. The swinging of a pendulum, the vibration of a guitar string.
Simple Harmonic Motion (SHM)
The simplest form of oscillatory motion in which the acceleration (and hence the force) is directly proportional to the displacement from the mean position and is always directed towards the mean position.
e.g. The motion of a block attached to a spring on a frictionless surface.
Period (T)
The smallest interval of time after which the motion is repeated. Its SI unit is second.
e.g. The time taken for one complete oscillation of a pendulum.
Frequency (ν)
The number of repetitions (oscillations) that occur per unit time. Its SI unit is hertz (Hz), where 1 Hz = 1 s⁻¹.
e.g. A heart beating 75 times per minute has a frequency of 1.25 Hz.
Displacement (x)
In the context of oscillations, it refers to the change in the physical property (e.g., position, angle, voltage) from its equilibrium value as a function of time.
e.g. The distance of a block from its equilibrium position on a spring, or the angular displacement of a pendulum from the vertical.
Amplitude (A)
The magnitude of the maximum displacement of the particle from its mean position.
e.g. If a block oscillates between +10 cm and -10 cm, its amplitude is 10 cm.
Phase (ωt + φ)
The time-dependent quantity (ωt + φ) in the argument of the sine or cosine function that determines the state of motion (position and velocity) of the particle at any time t.
e.g. For x = A cos(ωt + φ), the phase at time t is (ωt + φ).
Phase constant (φ)
The value of the phase at t = 0. It determines the initial position and velocity of the oscillating particle.
e.g. If φ = 0, the particle starts at x = A. If φ = π/2, the particle starts at x = 0.
Angular frequency (ω)
A measure of how rapidly the phase changes with time. It is related to the period and frequency by ω = 2π/T = 2πν. Its SI unit is radian per second.
e.g. For a body with a period of 1 s, ω = 2π rad s⁻¹.
Restoring force
The force acting on an oscillating body that is always directed towards the mean position (equilibrium) and tries to bring the body back to it.
e.g. The force exerted by a spring (F = -kx) or the tangential component of gravity on a pendulum (mg sin θ).
Linear harmonic oscillator
A particle oscillating under a force that is linearly proportional to its displacement (F = -kx).
e.g. A mass attached to a spring obeying Hooke's law.
Reference circle
A circle of radius equal to the amplitude (A) on which a reference particle (P) moves with uniform angular speed (ω). The projection of this particle on a diameter of the circle executes SHM.
e.g. The shadow of a ball moving in a horizontal circle on a vertical wall appears to move in SHM.
Formulas & Equations (13)
ν = 1/T
Reads as: Frequency equals one over period
Variables: ν = frequency (Hz), T = time period (s)
Use when: To find frequency when period is known, or vice-versa.
⚠ Students often forget that T must be in seconds. If T is given in minutes, they must convert to seconds first.
x(t) = A cos(ωt + φ)
Reads as: Displacement as a function of time equals amplitude times cosine of (omega t plus phi)
Variables: x(t) = displacement at time t (m), A = amplitude (m), ω = angular frequency (rad/s), t = time (s), φ = phase constant (rad)
Use when: To find the displacement of a particle in SHM at any instant of time.
⚠ Using degrees instead of radians in the argument of the cosine function. The argument (ωt + φ) is always in radians.
ω = 2π/T = 2πν
Reads as: Angular frequency equals two pi over period, or two pi times frequency
Variables: ω = angular frequency (rad/s), T = time period (s), ν = frequency (Hz)
Use when: To relate angular frequency to period or frequency.
⚠ Confusing angular frequency (ω) with ordinary frequency (ν). ω is 2π times larger than ν.
v(t) = -ωA sin(ωt + φ)
Reads as: Velocity as a function of time equals negative omega A times sine of (omega t plus phi)
Variables: v(t) = velocity at time t (m/s), ω = angular frequency (rad/s), A = amplitude (m), t = time (s), φ = phase constant (rad)
Use when: To find the instantaneous velocity of a particle in SHM.
⚠ Forgetting the negative sign, which indicates the direction of velocity relative to the positive x-axis.
a(t) = -ω²A cos(ωt + φ) = -ω²x(t)
Reads as: Acceleration as a function of time equals negative omega squared A times cosine of (omega t plus phi), which equals negative omega squared times displacement
Variables: a(t) = acceleration at time t (m/s²), ω = angular frequency (rad/s), A = amplitude (m), x(t) = displacement at time t (m)
Use when: To find the instantaneous acceleration of a particle in SHM. This is the key equation showing acceleration is proportional to displacement.
⚠ Writing a = +ω²x. The negative sign is crucial as it shows acceleration is always directed opposite to displacement (towards the mean position).
F(t) = -kx(t)
Reads as: Force as a function of time equals negative k times displacement
Variables: F(t) = restoring force (N), k = force constant (N/m), x(t) = displacement (m)
Use when: To find the force acting on a particle in SHM, or to identify if a motion is SHM (if F ∝ -x).
⚠ Forgetting the negative sign. The force is always opposite to the direction of displacement.
k = mω²
Reads as: Force constant equals mass times angular frequency squared
Variables: k = spring constant (N/m), m = mass (kg), ω = angular frequency (rad/s)
Use when: To find the spring constant if mass and angular frequency are known, or to find ω from k and m.
⚠ Confusing this with F = ma. This is a specific relation for SHM.
ω = √(k/m)
Reads as: Angular frequency equals the square root of k over m
Variables: ω = angular frequency (rad/s), k = spring constant (N/m), m = mass (kg)
Use when: To find the angular frequency of a spring-mass system.
⚠ Forgetting to take the square root.
T = 2π√(m/k)
Reads as: Period equals two pi times the square root of m over k
Variables: T = time period (s), m = mass (kg), k = spring constant (N/m)
Use when: To find the time period of a spring-mass system executing SHM.
⚠ Writing T = 2π√(k/m). The mass is in the numerator, and the spring constant is in the denominator.
K = ½ m v² = ½ m ω² A² sin²(ωt + φ) = ½ k A² sin²(ωt + φ)
Reads as: Kinetic energy equals one-half m v squared, which equals one-half m omega squared A squared sine squared of (omega t plus phi), which equals one-half k A squared sine squared of (omega t plus phi)
Variables: K = kinetic energy (J), m = mass (kg), v = velocity (m/s), ω = angular frequency (rad/s), A = amplitude (m), k = spring constant (N/m)
Use when: To find the kinetic energy of a particle in SHM at any instant.
⚠ Forgetting that the maximum kinetic energy is ½ kA², which occurs at the mean position (x=0).
U = ½ k x² = ½ k A² cos²(ωt + φ)
Reads as: Potential energy equals one-half k x squared, which equals one-half k A squared cosine squared of (omega t plus phi)
Variables: U = potential energy (J), k = spring constant (N/m), x = displacement (m), A = amplitude (m)
Use when: To find the potential energy of a particle in SHM at any instant.
⚠ Forgetting that the maximum potential energy is ½ kA², which occurs at the extreme positions (x = ±A).
E = K + U = ½ k A²
Reads as: Total mechanical energy equals kinetic energy plus potential energy, which equals one-half k A squared
Variables: E = total energy (J), K = kinetic energy (J), U = potential energy (J), k = spring constant (N/m), A = amplitude (m)
Use when: To find the total energy of a simple harmonic oscillator, which is constant.
⚠ Thinking total energy changes with time. It is constant for an undamped oscillator.
T = 2π√(L/g)
Reads as: Period equals two pi times the square root of L over g
Variables: T = time period (s), L = length of the pendulum (m), g = acceleration due to gravity (m/s²)
Use when: To find the time period of a simple pendulum for small angular displacements.
⚠ Forgetting that this formula is valid only for small oscillations (θ < 20°). Also, confusing L with the distance from the support to the centre of mass of the bob.
How It Works
Derivation of SHM from Uniform Circular Motion
  1. Consider a particle P moving uniformly on a circle of radius A with angular speed ω.
  2. At t = 0, let the position vector OP make an angle φ with the positive x-axis.
  3. At time t, the angle made by OP with the x-axis is (ωt + φ).
  4. The projection of OP on the x-axis is the point P'. The x-coordinate of P' is given by x(t) = A cos(ωt + φ).
  5. This equation defines SHM. Thus, the projection of uniform circular motion on a diameter of the circle executes SHM.
💡 The reference particle P moves with constant speed, but its projection P' does not. The force on P is centripetal, while the force on P' is the restoring force of SHM. They are different forces.
Derivation of Velocity and Acceleration in SHM
  1. Start with the displacement equation: x(t) = A cos(ωt + φ).
  2. Differentiate x(t) with respect to time to get velocity: v(t) = dx/dt = -ωA sin(ωt + φ).
  3. Differentiate v(t) with respect to time to get acceleration: a(t) = dv/dt = -ω²A cos(ωt + φ) = -ω²x(t).
  4. Alternatively, from the reference circle: the velocity of P is ωA (tangential). Its projection on the x-axis gives v(t) = -ωA sin(ωt + φ).
  5. The centripetal acceleration of P is ω²A (towards centre). Its projection on the x-axis gives a(t) = -ω²A cos(ωt + φ) = -ω²x(t).
💡 The negative sign in acceleration is critical; it shows that acceleration is always opposite to displacement, i.e., directed towards the mean position.
Derivation of Force Law for SHM
  1. From Newton's second law, F = ma.
  2. Substitute the expression for acceleration in SHM: a = -ω²x.
  3. Therefore, F = m(-ω²x) = -mω²x.
  4. Define the force constant k = mω², so F = -kx.
  5. This shows that the force is proportional to displacement and opposite in direction, which is the defining characteristic of SHM.
💡 The force law F = -kx is both a consequence and a definition of SHM. If a system obeys this law, its motion is SHM.
Energy Conservation in SHM
  1. The kinetic energy is K = ½ m v² = ½ m ω² A² sin²(ωt + φ).
  2. The potential energy is U = ½ k x² = ½ k A² cos²(ωt + φ).
  3. Since k = mω², the total energy E = K + U = ½ k A² [sin²(ωt + φ) + cos²(ωt + φ)].
  4. Using the identity sin²θ + cos²θ = 1, we get E = ½ k A².
  5. This shows that the total mechanical energy is constant and independent of time, confirming energy conservation for a conservative system.
💡 Energy is continuously exchanged between kinetic and potential forms. K is max at mean position (x=0), U is max at extremes (x=±A). The total energy is always ½ kA².
Derivation of Time Period of a Simple Pendulum
  1. Consider a simple pendulum of length L with a bob of mass m.
  2. The forces on the bob are tension (T) and weight (mg).
  3. Resolve mg into radial (mg cosθ) and tangential (mg sinθ) components.
  4. The tangential component mg sinθ provides the restoring torque about the support: τ = -L (mg sinθ).
  5. For small θ, sinθ ≈ θ (in radians), so τ ≈ -mgL θ.
  6. Using Newton's law for rotational motion, τ = I α, where I = mL² is the moment of inertia.
  7. Thus, mL² α = -mgL θ, which simplifies to α = -(g/L) θ.
  8. Comparing with the SHM equation a = -ω²x, we get ω² = g/L, so ω = √(g/L).
  9. Therefore, the time period is T = 2π/ω = 2π√(L/g).
💡 This derivation is valid only for small angular displacements (θ < 20°). The period is independent of the mass of the bob and the amplitude of oscillation.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Period and Frequency
✗ The period is the number of oscillations per second.
→ This is the definition of frequency. Period is the time taken for one oscillation.
✓ The period is the time taken for one complete oscillation, while frequency is the number of oscillations per unit time.
⚠ Forgetting the Negative Sign in Force and Acceleration
✗ In SHM, the force is F = kx.
→ The negative sign is crucial. It indicates the force is a restoring force, always directed opposite to the displacement.
✓ In SHM, the restoring force is F = -kx, where k is the force constant.
⚠ Using Degrees in Trigonometric Functions
✗ x = A cos(ωt + φ) where φ = 90°.
→ The argument of sine and cosine in SHM equations must be in radians, not degrees. 90° is π/2 radians.
✓ x = A cos(ωt + φ) where φ = π/2 rad.
⚠ Mixing Up the Formula for Period of a Spring and a Pendulum
✗ The period of a spring-mass system is T = 2π√(k/m).
→ The formula is T = 2π√(m/k). The mass is in the numerator, and the spring constant is in the denominator.
✓ The period of a spring-mass system is T = 2π√(m/k).
⚠ Thinking Total Energy Varies with Time
✗ The total energy of a simple harmonic oscillator is ½ kA² cos²(ωt + φ).
→ Total energy is constant. The expression given is for potential energy, which varies with time.
✓ The total energy of a simple harmonic oscillator is constant and equals ½ kA².
⚠ Assuming All Periodic Motion is SHM
✗ The motion of a ball bouncing off the ground is simple harmonic.
→ While the motion is periodic, it is not SHM because the restoring force is not proportional to displacement. The acceleration is constant (g) between bounces.
✓ The motion of a ball bouncing is periodic but not simple harmonic. SHM requires a restoring force proportional to displacement (F ∝ -x).
Compare & Contrast
Definition and Scope
Periodic Motion
Oscillatory Motion
Periodic motion repeats itself at regular intervals of time. Oscillatory motion is a specific type of periodic motion where an object moves to and fro about a mean position. All oscillatory motions are periodic, but not all periodic motions are oscillatory (e.g., uniform circular motion is periodic but not oscillatory).
Nature of Restoring Force
Simple Harmonic Motion (SHM)
Periodic but Non-SHM Motion
In SHM, the restoring force is directly proportional to the displacement from the mean position and is always directed towards it (F = -kx). In periodic but non-SHM motion, the restoring force is not proportional to displacement (e.g., the force in a bouncing ball is constant, not proportional to height).
Exam Focus
Marks pattern: 1 MCQ (1 mark), 1 Very Short Answer (1 mark), 1 Short Answer (3 marks), 1 Long Answer (5 marks). Total: 10 marks.
Most asked: Derivation of time period of a simple pendulum · Energy conservation in SHM (kinetic, potential, total energy) · Force law for SHM (F = -kx) and its relation to acceleration · Phase and phase constant problems · Velocity and acceleration as functions of time in SHM · Identifying SHM from given equations or force laws
Don't skip: Derivation of time period of a simple pendulum (5 marks); Numerical problems involving energy, velocity, and acceleration in SHM (3-5 marks); Problems on finding amplitude, phase constant, and displacement from initial conditions (3 marks)
One-Line Revision (read aloud before the exam)
  1. Periodic motion repeats after a fixed time interval T.
  2. Oscillatory motion is a to-and-fro periodic motion about a mean position.
  3. Simple Harmonic Motion (SHM) is the simplest oscillation where F = -kx.
  4. Displacement in SHM is given by x(t) = A cos(ωt + φ).
  5. Amplitude A is the maximum displacement from the mean position.
  6. Angular frequency ω = 2π/T = 2πν.
  7. Phase (ωt + φ) determines the state of motion at time t.
  8. Phase constant φ is the phase at t = 0.
  9. Velocity in SHM is v(t) = -ωA sin(ωt + φ).
  10. Acceleration in SHM is a(t) = -ω²x(t).
  11. The restoring force in SHM is F = -kx, where k = mω².
  12. Total energy in SHM is constant: E = ½ kA².
  13. Kinetic energy is maximum at the mean position (x=0).
  14. Potential energy is maximum at the extreme positions (x=±A).
  15. The period of a spring-mass system is T = 2π√(m/k).
  16. The period of a simple pendulum is T = 2π√(L/g) for small oscillations.
  17. The projection of uniform circular motion on a diameter is SHM.
  18. A simple pendulum's period is independent of the mass of the bob.
  19. For small angles, sinθ ≈ θ (in radians).
  20. Every oscillatory motion is periodic, but not every periodic motion is oscillatory.

MCQs — 10 Questions with Answers

Q1. A body executing periodic motion has a period of 2 seconds. What is its frequency?

  • A. 0.5 Hz ✓
  • B. 2 Hz
  • C. 4 Hz
  • D. 0.25 Hz

Answer: A — Using ν = 1/T, frequency = 1/2 = 0.5 Hz.

Q2. Which of the following is an example of oscillatory motion?

  • A. Earth orbiting the Sun
  • B. A ball rolling down an incline
  • C. A pendulum swinging back and forth ✓
  • D. A car moving at constant velocity

Answer: C — Only the pendulum moves to and fro about an equilibrium position; others are either circular or non-repetitive.

Q3. A human heartbeat has a frequency of 1.25 Hz. What is the period between consecutive beats?

  • A. 0.5 s
  • B. 0.8 s ✓
  • C. 1.25 s
  • D. 2.0 s

Answer: B — Period T = 1/ν = 1/1.25 = 0.8 s.

Q4. Which statement correctly distinguishes periodic motion from oscillatory motion?

  • A. Periodic motion must be oscillatory and vice versa
  • B. Oscillatory motion is to-and-fro about equilibrium; periodic motion just repeats after time T ✓
  • C. Oscillatory motion has no equilibrium position
  • D. Periodic motion only occurs in springs and pendulums

Answer: B — Circular motion is periodic but not oscillatory; oscillatory motion always involves to-and-fro motion about equilibrium.

Q5. A mass-spring system oscillates with amplitude 0.1 m. At the equilibrium position, which is true?

  • A. Velocity is zero and acceleration is maximum
  • B. Velocity is maximum and acceleration is zero ✓
  • C. Both velocity and acceleration are maximum
  • D. Both velocity and acceleration are zero

Answer: B — At equilibrium, the restoring force is zero so acceleration is zero, but velocity reaches its maximum value before the body returns.

Q6. The period of oscillation of a quartz crystal is 1 microsecond (10⁻⁶ s). Its frequency in Hz is:

  • A. 10⁶ Hz ✓
  • B. 10⁻⁶ Hz
  • C. 1 Hz
  • D. 10³ Hz

Answer: A — ν = 1/T = 1/(10⁻⁶) = 10⁶ Hz (1 megahertz).

Q7. Why is circular motion of a planet periodic but not oscillatory?

  • A. It repeats after a fixed period but does not move to and fro about an equilibrium position ✓
  • B. It does not have a defined period
  • C. It involves a restoring force directed toward center
  • D. Oscillatory and periodic motions are the same thing

Answer: A — Oscillatory motion requires to-and-fro motion about equilibrium; circular motion repeats but lacks this characteristic.

Q8. Which statement about damped oscillations is NOT correct?

  • A. Amplitude decreases with time due to friction
  • B. The oscillation eventually stops at equilibrium
  • C. Frequency increases as amplitude decreases ✓
  • D. Energy is dissipated through dissipative forces

Answer: C — In damped oscillations, frequency remains essentially constant while amplitude decreases; frequency does not increase.

Q9. A tuning fork vibrates 440 times per second. If you want to express this as a period, how many milliseconds would one complete vibration take?

  • A. 0.44 ms
  • B. 2.27 ms ✓
  • C. 440 ms
  • D. 4.4 ms

Answer: B — T = 1/ν = 1/440 s ≈ 0.00227 s = 2.27 ms (convert seconds to milliseconds).

Q10. Assertion: In oscillatory motion, the restoring force is always directed toward the equilibrium position. Reason: The force is proportional to displacement but opposite in sign.

  • A. Both Assertion and Reason are true; Reason explains Assertion ✓
  • B. Both are true; Reason does not explain Assertion
  • C. Assertion is true; Reason is false
  • D. Assertion is false; Reason is true

Answer: A — Both statements are correct and logically connected: F = −kx means force opposes displacement, always pulling toward equilibrium (F ∝ −x).

Flashcards

What is the key difference between periodic and oscillatory motion?

Periodic motion repeats after time T, but oscillatory motion repeats to and fro about an equilibrium position where a restoring force acts.

Define period (T) and write its SI unit.

Period is the smallest time interval after which motion repeats; SI unit is second (s).

Write the relationship between frequency ν and period T.

ν = 1/T, meaning frequency in hertz equals the reciprocal of period in seconds.

What is an equilibrium position in oscillatory motion?

The position where no net external force acts on the body; if disturbed, a restoring force brings it back.

Why is circular motion periodic but not oscillatory?

Circular motion repeats after time period but does not move to and fro about a mean position.

What physical property creates oscillations in a mass-spring system?

The restoring force provided by the spring, which is proportional to displacement and directed toward equilibrium.

Give two real-life examples where oscillatory motion is important.

Vibrating strings in musical instruments produce sound, and oscillating atoms in solids determine temperature through their average vibration energy.

What is meant by damping in oscillatory motion?

Damping is the gradual loss of oscillation amplitude due to friction and other dissipative forces.

In simple harmonic motion, what is the direction of the net force?

Always directed toward the mean (equilibrium) position, opposite to the direction of displacement.

How do oscillations and vibrations differ in terminology?

Oscillations are called when frequency is low (tree branches), vibrations when frequency is high (musical strings).

Important Board Questions

Define oscillatory motion. Give two real-life examples and explain why they are considered oscillatory. [2 marks]

State that oscillatory motion is periodic to-and-fro about equilibrium with restoring force. Pick examples like pendulum or mass-spring and mention the restoring force in each case.

A body executing periodic motion completes 50 oscillations in 10 seconds. Calculate its frequency and period. If the amplitude of oscillation is 5 cm, how would you distinguish this from circular motion? [5 marks]

Use ν = number of oscillations / time to find frequency; then T = 1/ν to find period. Explain that oscillatory motion is to-and-fro about equilibrium while circular motion repeats but does not oscillate about a mean position.

Explain why every oscillatory motion is periodic, but not every periodic motion is oscillatory. Use appropriate diagrams (height-time graphs) to illustrate your answer with at least two contrasting examples. [6 marks]

Define periodic (repeats after time T) and oscillatory (to-and-fro about equilibrium). Use graphs: smooth sinusoidal curve for true oscillation (pendulum), stepped parabolic curve for non-oscillatory periodic motion (bouncing ball), and circular path for periodic but non-oscillatory motion (planet orbit). Identify restoring force in oscillatory cases.

Assertion-Reason Questions for Oscillations

Q1. Assertion (A):

Every oscillatory motion is a periodic motion.

Reason (R):

Oscillatory motion involves to-and-fro motion about an equilibrium position with the same pattern repeating at regular time intervals.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. An oscillatory motion, by definition, involves repeated to-and-fro motion about an equilibrium, which means the motion pattern repeats at regular intervals of the period T. This repetition is precisely what makes it periodic. The reason correctly explains why every oscillatory motion must be periodic.

Q2. Assertion (A):

Circular motion of a planet is an example of periodic motion.

Reason (R):

The planet returns to the same position after a fixed time interval equal to its orbital period.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—planetary motion is periodic with period equal to one year. However, the reason, while true, does NOT correctly explain why it is periodic according to the definition given in the chapter. The chapter defines periodic motion as motion that 'repeats itself at regular intervals of time,' which includes the planet returning to the same position, but the chapter's deeper point is that planetary motion is NOT oscillatory (no to-and-fro about equilibrium), distinguishing it from true oscillatory periodic motion. The reason states a true fact but misses the nuance of the classification.

Q3. Assertion (A):

If the period of oscillation is halved, the frequency becomes twice the original frequency.

Reason (R):

Frequency is inversely proportional to period (ν = 1/T).

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason is the correct mathematical explanation. If T_new = T_old/2, then ν_new = 1/(T_old/2) = 2/T_old = 2ν_old. The inverse proportionality directly explains why halving the period doubles the frequency. This is a straightforward application of the fundamental relationship between period and frequency.

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Case-Based Questions (CBQ) for Oscillations

Case Study:

A wall clock with a pendulum is installed in a laboratory. The pendulum completes 60 oscillations in 120 seconds. A student observes the clock's behavior over one hour (3600 seconds) and notes that the amplitude gradually decreases due to air resistance and friction at the pivot. However, the clock mechanism provides a small energy input every few seconds to maintain approximately constant frequency. Despite the decreasing amplitude, the clock continues to keep accurate time.

Q1. What is the frequency of the pendulum in Hz?

  • A. 0.5 Hz ✓
  • B. 60 Hz
  • C. 120 Hz
  • D. 2 Hz

Answer: A

Q2. How many complete oscillations will the pendulum complete in one hour?

  • A. 1800 oscillations ✓
  • B. 3600 oscillations
  • C. 7200 oscillations
  • D. 180 oscillations

Answer: A

Q3. Why does the clock continue to keep accurate time despite decreasing amplitude?

  • A. The energy input increases the frequency
  • B. The friction force increases with time
  • C. Amplitude does not affect the period of oscillation ✓
  • D. Decreasing amplitude automatically increases frequency

Answer: C

Q4. What is the period of the pendulum?

  • A. 0.5 seconds
  • B. 2 seconds ✓
  • C. 60 seconds
  • D. 120 seconds

Answer: B

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