Chapter 13: Oscillations
Unit X – Oscillations and Waves
13.1 Periodic Motion
A motion that repeats itself at regular intervals of time is called periodic motion.
- Time period (T): Time for one complete oscillation.
- Frequency (f): Number of oscillations per second. \(f = 1/T\), unit: Hz.
- Angular frequency: \(\omega = 2\pi f = 2\pi/T\)
13.2 Simple Harmonic Motion (SHM)
SHM is a special periodic motion where the restoring force is proportional to displacement and directed towards the mean position.
\[ F = -kx \Rightarrow ma = -kx \Rightarrow a = -\omega^2 x \]
where \(\omega = \sqrt{k/m}\)
Displacement in SHM:
\[ x(t) = A\sin(\omega t + \phi) \]
or
\[ x(t) = A\cos(\omega t + \phi) \]
where:
- \(A\) = amplitude (maximum displacement)
- \(\omega\) = angular frequency
- \(\phi\) = initial phase (phase constant)
13.3 Velocity and Acceleration in SHM
Velocity:
\[ v = \frac{dx}{dt} = A\omega\cos(\omega t + \phi) \]
At displacement x: \(v = \omega\sqrt{A^2 - x^2}\)
- Maximum velocity \(v_{max} = A\omega\) (at mean position, \(x = 0\))
- Minimum velocity = 0 (at extreme positions, \(x = \pm A\))
Acceleration:
\[ a = -\omega^2 x \]
- Maximum acceleration \(a_{max} = \omega^2 A\) (at extreme positions)
- Acceleration = 0 at mean position
13.4 SHM Displacement Graph
13.5 Energy in SHM
Potential energy:
\[ U = \frac{1}{2}kx^2 = \frac{1}{2}m\omega^2 x^2 \]
Kinetic energy:
\[ KE = \frac{1}{2}m\omega^2(A^2 - x^2) \]
Total mechanical energy (constant):
\[ E = KE + U = \frac{1}{2}m\omega^2 A^2 = \frac{1}{2}kA^2 \]
13.6 Simple Pendulum
A simple pendulum of length \(L\) oscillates with period:
\[ T = 2\pi\sqrt{\frac{L}{g}} \]
- Valid for small angles (\(\theta \lesssim 10°\))
- Period is independent of mass and amplitude (for small oscillations)
- Frequency: \(f = \frac{1}{2\pi}\sqrt{\frac{g}{L}}\)
Factor affecting T:
- Increasing L → T increases
- Increasing g → T decreases (pendulum oscillates faster)
13.7 Spring-Mass System
For a mass \(m\) on spring of constant \(k\):
Horizontal: \[ T = 2\pi\sqrt{\frac{m}{k}} \]
Vertical (same formula — gravity only shifts equilibrium, doesn’t change T): \[ T = 2\pi\sqrt{\frac{m}{k}} \]
Springs in series: \[ \frac{1}{k_{eff}} = \frac{1}{k_1} + \frac{1}{k_2} \]
Springs in parallel: \[ k_{eff} = k_1 + k_2 \]
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(x = A\sin(\omega t + \phi)\) | SHM displacement |
| \(v_{max} = A\omega\) | Maximum velocity |
| \(a_{max} = A\omega^2\) | Maximum acceleration |
| \(E = \frac{1}{2}kA^2\) | Total energy in SHM |
| \(T = 2\pi\sqrt{L/g}\) | Simple pendulum period |
| \(T = 2\pi\sqrt{m/k}\) | Spring-mass period |
Practice Questions
Section A – MCQ (1 mark each)
Q1. In SHM, when displacement is half the amplitude, the ratio KE:PE is:
(a) 1:3 (b) 3:1 (c) 1:4 (d) 3:4
Answer
(b) 3:1
\(PE = \frac{1}{2}k(A/2)^2 = \frac{1}{4} \times \frac{1}{2}kA^2 = \frac{E}{4}\)
\(KE = E - PE = \frac{3E}{4}\)
Ratio KE:PE = 3:1
Q2. The time period of a simple pendulum is 2 s. If its length is quadrupled, the new time period is:
(a) 1 s (b) 2 s (c) 4 s (d) 8 s
Answer
(c) 4 s — \(T \propto \sqrt{L}\), so quadrupling L doubles T: \(2 \times 2 = 4 \text{ s}\).
Section B – Short Answer (2–3 marks)
Q3. A particle executes SHM with amplitude 10 cm and period 4 s. Find:
(i) Maximum velocity (ii) Maximum acceleration (iii) Velocity at displacement 6 cm
Answer
\(\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4} = \dfrac{\pi}{2} \text{ rad/s}\), \(A = 0.1 \text{ m}\)
(i) \(v_{max} = A\omega = 0.1 \times \dfrac{\pi}{2} \approx \mathbf{0.157 \text{ m/s}}\)
(ii) \(a_{max} = A\omega^2 = 0.1 \times \dfrac{\pi^2}{4} \approx \mathbf{0.247 \text{ m/s}^2}\)
(iii) \(v = \omega\sqrt{A^2 - x^2} = \dfrac{\pi}{2}\sqrt{(0.1)^2 - (0.06)^2} = \dfrac{\pi}{2}\sqrt{0.0064} = \dfrac{\pi}{2} \times 0.08 \approx \mathbf{0.126 \text{ m/s}}\)
Section D – Competency-Based Questions
Q4. (Case Study) A grandfather clock uses a pendulum of length 1 m. It shows correct time at sea level (\(g = 9.8 \text{ m/s}^2\)).
(i) Find the time period of the pendulum.
(ii) If the clock is taken to a mountain where \(g = 9.7 \text{ m/s}^2\), will it gain or lose time? By how much per day?
(iii) What should be the length of the pendulum to have \(T = 1\) s?
Answer
(i) \(T = 2\pi\sqrt{L/g} = 2\pi\sqrt{1/9.8} = 2\pi \times 0.319 \approx \mathbf{2.006 \text{ s}}\)
(ii) At lower g: \(T’ = 2\pi\sqrt{1/9.7} \approx 2.015 \text{ s}\) — period increases → clock runs slower → loses time.
Extra time per oscillation: \(\Delta T \approx 0.009 \text{ s}\)
Oscillations per day: \(\dfrac{86400}{2} = 43200\)
Time lost per day: \(43200 \times 0.009 \approx \mathbf{389 \text{ s} \approx 6.5 \text{ min}}\)
(iii) \(T = 2\pi\sqrt{L/g} = 1 \Rightarrow L = \dfrac{g}{4\pi^2} = \dfrac{9.8}{39.48} \approx \mathbf{0.248 \text{ m}}\)