What this chapter is about
Oscillations describe any motion that repeats itself in a regular pattern about a central, equilibrium position. A child on a swing, the bob of a pendulum, the vibrating string of a sitar, and the back-and-forth motion of atoms in a solid are all examples. This chapter introduces the physics behind such periodic motion, focusing on the simplest and most important case: simple harmonic motion (SHM).
A Class 11 student meets oscillations now because the ideas here connect mechanics (force, acceleration, energy) with waves, sound and even alternating-current electricity studied later. Understanding SHM gives you a powerful template: once you recognise that the restoring force on an object is proportional to its displacement, you can predict its frequency, energy and phase without solving complicated equations each time.
After working through this chapter you should be able to identify whether a given motion is simple harmonic, write the displacement, velocity and acceleration as functions of time, calculate time period and frequency for a spring–mass system and a simple pendulum, and analyse energy transformations during oscillation.
Key ideas
- Periodic motion is any motion that repeats exactly after a fixed time interval called the time period T. The reciprocal 1/T is the frequency f (unit: hertz, Hz).
- Oscillatory (vibratory) motion is periodic motion where the object moves to and fro about a mean position; not every periodic motion is oscillatory (e.g. uniform circular motion is periodic but not oscillatory).
- Simple harmonic motion (SHM) is oscillatory motion in which the restoring force F is directly proportional to the displacement x from equilibrium and directed opposite to it: F = −kx, where k is a positive constant.
- In SHM the displacement varies sinusoidally with time: x(t) = A cos(ωt + φ), where A is the amplitude, ω = 2πf is the angular frequency (rad/s), and φ is the initial phase.
- Velocity in SHM is v = −Aω sin(ωt + φ); it is maximum at the mean position and zero at the extreme positions.
- Acceleration in SHM is a = −ω²x; it is maximum at the extremes (magnitude Aω²) and zero at the mean position.
- Total mechanical energy E = (1/2)kA² remains constant during SHM; kinetic and potential energies interchange but their sum does not change (in the absence of damping).
- For a mass m on a spring of constant k, the time period T = 2π√(m/k). For a simple pendulum of length L (small-angle oscillations), T = 2π√(L/g), where g is acceleration due to gravity.
Formulas and facts to remember
- Expression: T = 1/f · Meaning: Time period (s) equals reciprocal of frequency (Hz).
- Expression: ω = 2πf = 2π/T · Meaning: Angular frequency in rad/s.
- Expression: x = A cos(ωt + φ) · Meaning: Displacement in SHM; A is amplitude, φ is phase constant.
- Expression: v = dx/dt = −Aω sin(ωt + φ) · Meaning: Velocity in SHM; maximum value v_max = Aω.
- Expression: a = dv/dt = −ω²x · Meaning: Acceleration in SHM; always opposite to displacement.
- Expression: T (spring) = 2π√(m/k) · Meaning: Time period of a mass–spring oscillator.
- Expression: T (pendulum) = 2π√(L/g) · Meaning: Time period of a simple pendulum (small angle).
- Expression: E = (1/2)kA² = (1/2)mω²A² · Meaning: Total mechanical energy in SHM (constant).
- Expression: KE = (1/2)mv² ; PE = (1/2)kx² · Meaning: Kinetic and potential energies; their sum equals E.
Worked examples
Example 1 – Spring–mass system
A 0.50 kg block attached to a horizontal spring oscillates with an amplitude of 0.10 m and completes 4 oscillations in 8.0 s. Find the spring constant.
Step 1. Time period T = 8.0 s / 4 = 2.0 s.
Step 2. Use T = 2π√(m/k). Rearranging gives k = 4π²m / T².
Step 3. Substitute values: k = 4 × (3.14)² × 0.50 / (2.0)² = 4 × 9.87 × 0.50 / 4 = 4.9 N/m (two significant figures).
The spring constant is approximately 4.9 N/m.
Example 2 – Simple pendulum on Earth
A simple pendulum has a length of 1.0 m. Calculate its time period and frequency (take g = 9.8 m/s²).
Step 1. T = 2π√(L/g) = 2 × 3.14 × √(1.0 / 9.8).
Step 2. √(1.0/9.8) = √0.102 ≈ 0.32 s.
Step 3. T ≈ 2 × 3.14 × 0.32 ≈ 2.0 s.
Step 4. Frequency f = 1/T = 1/2.0 = 0.50 Hz.
The pendulum takes about 2.0 s for one swing and oscillates at 0.50 Hz.
Example 3 – Energy at a given displacement
A particle executes SHM with amplitude 0.04 m and angular frequency 50 rad/s. Its mass is 0.20 kg. Find the kinetic energy when displacement is 0.02 m.
Step 1. Total energy E = (1/2)mω²A² = 0.5 × 0.20 × (50)² × (0.04)² = 0.5 × 0.20 × 2500 × 0.0016 = 0.40 J.
Step 2. Potential energy at x = 0.02 m: PE = (1/2)mω²x² = 0.5 × 0.20 × 2500 × (0.02)² = 0.10 J.
Step 3. Kinetic energy KE = E − PE = 0.40 − 0.10 = 0.30 J.
At a displacement of 0.02 m the kinetic energy is 0.30 J.
Common mistakes
- Assuming acceleration is maximum at the mean position → In SHM acceleration is zero at the mean position; it is maximum at the extremes where restoring force is greatest.
- Confusing frequency f (in Hz) with angular frequency ω (in rad/s) → Always check units; ω = 2πf.
- Using T = 2π√(L/g) for large swing angles → This formula holds only for small angles (typically less than about 15°). For larger angles, the motion is still periodic but not exactly SHM.
- Forgetting the negative sign in F = −kx or a = −ω²x → The minus sign shows the force or acceleration opposes the displacement; it is essential for SHM.
- Treating amplitude and displacement as the same quantity → Amplitude A is the maximum displacement; x changes with time while A stays constant.
Quick revision
- SHM occurs when the restoring force is proportional to displacement and opposite in direction: F = −kx.
- Displacement x = A cos(ωt + φ); velocity and acceleration follow by differentiation.
- Time period for a spring: T = 2π√(m/k); for a simple pendulum: T = 2π√(L/g).
- Total mechanical energy (1/2)kA² stays constant; KE and PE exchange during each cycle.
- At the mean position: velocity is maximum, acceleration is zero. At extreme positions: velocity is zero, acceleration is maximum.