A swing, a guitar string and a mass on a spring all move back and forth. Not all such motion is "simple harmonic". The test is one line: is the pull back towards the middle proportional to how far the body has moved? This note gives the words, the test and the spring example.
Chapter 14 of the PECTAA Physics 12 book is Simple Harmonic Motion. It opens with oscillatory motion (14.1), then defines simple harmonic motion (14.2) and studies practical systems (14.3). This note is for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.
What to know
Oscillatory motion. A to-and-fro motion about a mean position. It is periodic: it repeats after equal time intervals. The book's everyday examples include a simple pendulum, a mass on a spring, a tuning fork, atoms in a solid, a swing, air molecules as sound passes, guitar and violin strings, and the wings of a flying bird.
Every such body has a mean (equilibrium) position. When it is pushed away, a restoring force pulls it back, and that keeps it oscillating.
Terms to learn.
- Vibration: one complete round trip of the body about its mean position, from one extreme to the other and back.
- Instantaneous displacement (x): the distance of the body from its mean position at any moment.
- Amplitude (x₀): the largest displacement on either side of the mean position.
- Time period (T): the time for one vibration. Unit: second.
- Frequency (f): the number of vibrations per second. f = 1/T. Unit: hertz (Hz), one oscillation per second. Dimension [T⁻¹].
- Angular frequency (ω): angular displacement per unit time. One cycle is 2π radians in time T, so ω = 2π/T = 2πf. Unit: rad s⁻¹.
Simple harmonic motion (14.2). An oscillation produced by a restoring force with these conditions:
- The restoring force is directly proportional to the displacement from the mean position.
- Force and displacement follow Hooke's law, F = −kx, where k is the spring constant and depends on the spring's material.
- The acceleration is proportional to the displacement, a ∝ −x. The minus sign means the acceleration always points towards the mean position.
The book's side note warns that not every periodic vibration is simple harmonic, because not every restoring force is proportional to displacement.
Mass on a spring (14.3). A mass m sits on a smooth horizontal surface, joined to a spring of constant k. Pull it a distance x and let go. The spring's restoring force is −kx. By Newton's second law, F = ma. Equating the two:
ma = −kx, so a = −(k/m)x
Because k/m is constant, a ∝ −x. This is the mathematical form of simple harmonic motion. Comparing with a = −ω²x gives ω = √(k/m), so the time period is T = 2π√(m/k).
The mass passes the mean position because of its inertia, compresses the spring on the other side, and keeps oscillating between the two extremes.
Key points
- SHM test: a ∝ −x. If the acceleration is not proportional to displacement, it is not SHM.
- Three linked quantities: T, f and ω. Know one and you know all three.
- Spring period: a heavier mass gives a longer period; a stiffer spring gives a shorter one.
Worked examples
Example 1 (book example). A particle in SHM has angular frequency 2 rad s⁻¹. Find its frequency and time period.
- ω = 2πf, so f = 2 / 2π = 1/π ≈ 0.318 Hz.
- T = 1/f = π ≈ 3.14 s.
Example 2 (book self-check style). A body on a spring makes 10 vibrations in 20 seconds. Find T and f.
- T = 20 / 10 = 2 s.
- f = 1/T = 0.5 Hz.
Example 3. A 0.5 kg mass is on a spring with k = 50 N m⁻¹. Find ω and T.
- ω = √(k/m) = √(50 / 0.5) = √100 = 10 rad s⁻¹.
- T = 2π/ω = 2π/10 ≈ 0.628 s.
Example 4. For that mass, find the acceleration when x = 0.04 m.
- a = −(k/m)x = −100 × 0.04 = −4 m s⁻².
- The minus sign shows it points back towards the mean position.
Common mistakes in the exam
- Calling every oscillation SHM. It is SHM only when the restoring force is proportional to displacement.
- Dropping the minus sign in a = −(k/m)x. The sign shows the direction towards the mean position.
- Mixing f and ω. ω = 2πf, not f.
- Taking amplitude as the full swing. Amplitude is measured from the mean position to one extreme.
- Counting half a swing as one vibration. One vibration is a full round trip.
Exam technique
- In definitions, use the words "mean position", "restoring force" and "proportional to displacement".
- For the spring derivation, write Hooke's law, then Newton's second law, then equate. Three clear lines earn the marks.
- Show units on every answer: s, Hz, rad s⁻¹, m s⁻².
- Check your board's notice for the topics in your paper.
Practice questions
- Define amplitude. Answer: The largest displacement of the vibrating body on either side of its mean position.
- A pendulum completes 30 vibrations in 60 s. Find its frequency. Answer: T = 2 s, so f = 0.5 Hz.
- Find ω for a body with T = 0.5 s. Answer: ω = 2π/0.5 = 4π ≈ 12.57 rad s⁻¹.
- A 2 kg mass on a spring of k = 200 N m⁻¹. Find ω. Answer: ω = √(200/2) = 10 rad s⁻¹.
- Write the mathematical form of SHM. Answer: a ∝ −x, or a = −(k/m)x for a mass on a spring.
Ready to practise this? Continue in 2nd Year Physics.