How to Use This Page
1. Solve first, check later — Attempt every exercise question on your own before reading the solutions below. 2. Follow the sequence — Work through sections I to VI in order; later sections build on earlier ones. 3. Understand the method — Focus on the reasoning and formula selection, not just the final answer. 4. Practise numericals — Re-work the worked examples with different values until you can set them up without hints. 5. Review mistakes — Read the "Common Mistakes to Avoid" section before every exam.
Key Concepts (Quick Recap)
SHM Definitions and Equations
| Quantity | Formula | SI Unit |
|---|---|---|
| Displacement | x = A sin(ωt + φ) | m |
| Velocity | v = Aω cos(ωt + φ) = ω√(A² − x²) | m/s |
| Maximum velocity | v_max = Aω | m/s |
| Acceleration | a = −ω²x = −Aω² sin(ωt + φ) | m/s² |
| Maximum acceleration | a_max = Aω² | m/s² |
| Angular frequency | ω = 2π/T = 2πf | rad/s |
| Frequency | f = 1/T | Hz |
| Time period | T = 2π/ω = 1/f | s |
Restoring Force and Spring Constant
| Quantity | Formula |
|---|---|
| Restoring force | F = −kx (Hooke's law for SHM) |
| Spring constant | k = F/x |
| Angular frequency (spring) | ω = √(k/m) |
| Time period (spring) | T = 2π√(m/k) |
Simple Pendulum
| Quantity | Formula |
|---|---|
| Restoring force | F = −mg sinθ ≈ −mgθ (small angle) |
| Time period | T = 2π√(L/g) |
| Angular frequency | ω = √(g/L) |
| Seconds pendulum | L ≈ 0.9936 m (T = 2 s) |
Energy in SHM
| Quantity | Formula |
|---|---|
| Kinetic energy | KE = (1/2)mv² = (1/2)mω²(A² − x²) |
| Potential energy | PE = (1/2)kx² = (1/2)mω²x² |
| Total energy | E = KE + PE = (1/2)mω²A² = (1/2)kA² |
| KE at mean position | KE_max = (1/2)mω²A² |
| PE at extreme position | PE_max = (1/2)kA² = (1/2)mω²A² |
Types of Oscillations
| Type | Description | Amplitude |
|---|---|---|
| Free | No external force, no damping | Constant |
| Damped | Resistive force opposes motion | Decreases exponentially |
| Forced | External periodic force drives the system | Depends on driving frequency |
| Resonance | Driving frequency = natural frequency | Maximum |
Damped Oscillations
| Quantity | Formula |
|---|---|
| Damping force | F_d = −bv (viscous damping) |
| Equation of motion | mẍ + bẋ + kx = 0 |
| Damped frequency | ω' = √(ω₀² − b²/4m²) |
| Amplitude decay | A(t) = A₀ e^(−bt/2m) |
I. Choose the Best Answer (MCQs)
1. The time period of a simple pendulum of length 1 m on Earth (g = 9.8 m/s²) is approximately:
- (a) 1.0 s
- (b) 2.0 s
- (c) 3.14 s
- (d) 0.5 s
2. In SHM, the acceleration is proportional to:
- (a) Velocity
- (b) Displacement from equilibrium
- (c) Square of displacement
- (d) Time period
3. The total energy of a particle in SHM is proportional to:
- (a) Amplitude (A)
- (b) Square of amplitude (A²)
- (c) Square root of amplitude
- (d) Inverse of amplitude
4. A spring-mass system oscillates with a period of 2 s. If the mass is quadrupled, the new period is:
- (a) 1 s
- (b) 2 s
- (c) 4 s
- (d) 8 s
5. At the mean position in SHM:
- (a) KE is maximum, PE is maximum
- (b) KE is minimum, PE is maximum
- (c) KE is maximum, PE is minimum
- (d) Both KE and PE are minimum
6. The angular frequency of a spring-mass system with k = 100 N/m and m = 0.25 kg is:
- (a) 10 rad/s
- (b) 20 rad/s
- (c) 50 rad/s
- (d) 400 rad/s
7. Resonance occurs when the driving frequency equals:
- (a) Zero
- (b) Twice the natural frequency
- (c) The natural frequency
- (d) Half the natural frequency
II. Fill in the Blanks
1. The time period of a simple pendulum depends on its length and acceleration due to gravity, but not on its mass. 2. In SHM, the restoring force is proportional to displacement from the equilibrium position. 3. The total energy of a particle in SHM remains constant throughout the motion. 4. The phase difference between displacement and velocity in SHM is π/2 (or 90°). 5. At the extreme position in SHM, the kinetic energy is zero. 6. The time period of a spring-mass system is T = 2π√(m/k). 7. A pendulum with a time period of 2 seconds is called a seconds pendulum. 8. In damped oscillations, the amplitude decreases exponentially with time. 9. The SI unit of angular frequency is rad/s. 10. The relation between frequency f and angular frequency ω is ω = 2πf. 11. The maximum displacement from the mean position is called the amplitude. 12. When the driving frequency equals the natural frequency, the phenomenon is called resonance.
III. True or False (with corrections)
IV. Match the Following
| Column A | Column B |
|---|---|
| 1. Simple pendulum | (e) T = 2π√(L/g) |
| 2. Spring-mass system | (d) T = 2π√(m/k) |
| 3. Total energy in SHM | (a) (1/2)kA² |
| 4. Maximum velocity | (c) Aω |
| 5. Phase difference (displacement vs velocity) | (f) π/2 |
| 6. Resonance | (b) Driving frequency equals natural frequency |
V. Answer Briefly (2-Mark Questions)
VI. Answer in Detail (5-Mark Questions)
1. Derive the time period of a simple pendulum and solve numericals.
A simple pendulum consists of a point mass (bob) suspended from a fixed point by a massless, inextensible string of length L.
Derivation:
When the bob is displaced by angle θ from the vertical:
- Restoring force along the arc: F = −mg sinθ
- For small angles (θ < 10°), sinθ ≈ θ in radians
- Displacement along the arc: x = Lθ, so θ = x/L
Worked Example 1: A simple pendulum on Earth has a time period of 2 s. Find its length. What would be its time period on the Moon where g_moon = g/6?
Worked Example 2: A simple pendulum has a length of 2.5 m. Find its frequency and angular frequency. (g = 9.8 m/s²)
2. Derive expressions for KE, PE, and total energy in SHM and show they are constant. Solve a numerical.
For a particle in SHM: x = A sin(ωt + φ)
Velocity: v = dx/dt = Aω cos(ωt + φ) At any displacement x: v = ω√(A² − x²)
Kinetic Energy:
Potential Energy:
Total Energy:
Energy variation with displacement:
| x | KE | PE | E |
|---|---|---|---|
| 0 (mean) | (1/2)mω²A² (max) | 0 | (1/2)mω²A² |
| ±A/√2 | (1/2)mω²A²/2 | (1/2)mω²A²/2 | (1/2)mω²A² |
| ±A (extreme) | 0 | (1/2)mω²A² (max) | (1/2)mω²A² |
Worked Example: A body of mass 0.5 kg executes SHM with amplitude 0.2 m and frequency 5 Hz. Find the maximum velocity, maximum acceleration, maximum KE, and total energy.
3. Derive the time period of a spring-mass system for horizontal and vertical oscillations, and compare with a simple pendulum.
Horizontal spring-mass system:
When a mass m is attached to a spring of constant k and displaced by x from equilibrium:
- Restoring force: F = −kx (Hooke's law)
- By Newton's second law: ma = −kx
- a = −(k/m)x
Vertical spring-mass system:
When a mass m hangs vertically from a spring:
- At equilibrium, the spring extends by δ: kδ = mg, so δ = mg/k
- If displaced by y from the new equilibrium: F_net = −k(δ + y) + mg = −ky
Comparison with simple pendulum:
| Property | Spring-mass system | Simple pendulum |
|---|---|---|
| Restoring force | F = −kx | F = −mg sinθ ≈ −mgx/L |
| Time period | T = 2π√(m/k) | T = 2π√(L/g) |
| Depends on mass? | Yes (T ∝ √m) | No |
| Depends on gravity? | No (horizontal) | Yes (T ∝ 1/√g) |
| Amplitude independence | Yes (isochronous) | Yes (small angles) |
Worked Example: A spring of force constant k = 200 N/m has a mass of 0.5 kg attached to it. Find the time period and frequency. If the same spring-mass system is taken to the Moon (g_moon = 1.63 m/s²), what changes?
Common Mistakes to Avoid
1. Confusing angular frequency (ω) with ordinary frequency (f) — Angular frequency ω = 2πf is measured in rad/s, while ordinary frequency f = 1/T is measured in Hz. Always check which one a question asks for and convert using ω = 2πf.
2. Using pendulum formula for spring-mass system or vice versa — T = 2π√(L/g) is for a simple pendulum; T = 2π√(m/k) is for a spring-mass system. These are fundamentally different systems with different time periods.
3. Forgetting the negative sign in SHM equations — The restoring force is F = −kx and acceleration is a = −ω²x. The negative sign indicates that force/acceleration is always directed toward the equilibrium position. Dropping it changes the physics.
4. Assuming pendulum period depends on mass or amplitude — The time period of a simple pendulum depends only on L and g (T = 2π√(L/g)). It does not depend on the mass of the bob or the amplitude (for small angles). This is the law of isochronism.
5. Mixing up conditions at mean and extreme positions — At the mean position: v = maximum, a = 0, KE = maximum, PE = 0. At the extreme position: v = 0, a = maximum, KE = 0, PE = maximum. Getting these swapped is a very common error.
6. Using degrees instead of radians — In SHM equations, the phase angle (ωt + φ) must be in radians, not degrees. When computing sin(ωt + φ), convert degrees to radians first. The formula T = 2π√(L/g) uses radians inherently.
7. Confusing damped frequency with natural frequency — Damped oscillations have a frequency ω' = √(ω₀² − b²/4m²), which is less than the natural frequency ω₀. The damping reduces both the amplitude and the frequency. Do not use ω₀ in place of ω' for damped systems.
8. Not using small angle approximation for pendulum — The formula T = 2π√(L/g) is valid only for small angles (θ < 10°). For larger angles, the period is longer than this formula predicts. If a question specifies a large angle, mention that the small-angle approximation may not hold.
Exam Checklist
- [ ] I can define SHM and write the differential equation d²x/dt² + ω²x = 0
- [ ] I can derive and use displacement, velocity, and acceleration equations for SHM
- [ ] I can derive and apply T = 2π√(L/g) for a simple pendulum
- [ ] I can derive and apply T = 2π√(m/k) for a spring-mass system
- [ ] I can calculate KE, PE, and total energy at any displacement in SHM
- [ ] I can show that total energy is constant in SHM (E = (1/2)mω²A²)
- [ ] I can identify conditions at mean and extreme positions (v, a, KE, PE)
- [ ] I can distinguish between free, damped, and forced oscillations and define resonance
- [ ] I can convert between frequency (f), angular frequency (ω), and time period (T)
- [ ] I can solve numericals involving pendulum, spring-mass, and energy in SHM