Spring force and force constant
Stretch or squeeze a spring by x and it pulls or pushes back with F = −kx (Hooke’s law, for small x). The minus sign means the force is opposite to the stretch.
k is the force constant or spring constant: the force needed per metre of stretch. Unit: N/m. A big k means a stiff spring.
Newton’s second law: ma = −kx, so a = −(k/m) x. This is SHM with ω² = k/m.
T = 2π√(m/k), f = (1/2π)√(k/m).
Springs joined together
- Series (one after the other): 1/k = 1/k₁ + 1/k₂ (softer).
- Parallel (side by side): k = k₁ + k₂ (stiffer).
- Cutting a spring into n equal pieces makes each piece n times stiffer (k′ = nk).
Vertical spring
Hang a mass on a vertical spring. It stretches by d until kd = mg, so k = mg/d. About this new resting point it still does SHM with the same T = 2π√(m/k). Gravity only shifts the middle; it does not change the period. Also T = 2π√(d/g).
Simple pendulum and its time period
A simple pendulum is a small heavy bob hung on a light string that does not stretch, of length L (measured to the centre of the bob).
Pull it aside by angle θ. Gravity mg has a part mg sinθ along the arc, pulling it back. For small angles (less than about 10°), sinθ ≈ θ = x/L, so
F ≈ −(mg/L) x → a = −(g/L) x.
That is SHM with ω² = g/L, so T = 2π√(L/g).
- T does not depend on the mass of the bob (mass cancels).
- T does not depend on the amplitude for small swings (isochronism).
- T ∝ √L and T ∝ 1/√g.
A seconds pendulum has T = 2 s; on Earth its length is about 0.99 m.
Try it: find g at home
Hang a small heavy object on a thread 1.00 m long. Pull it aside a little (about the width of your hand) and time 20 swings. Suppose you get 40.1 s, so T = 2.005 s. Then g = 4π²L/T² = 4 × 9.87 × 1.00 / 4.02 ≈ 9.8 m/s². Now repeat with a 0.25 m thread: predict T first (half of 2.0 s), then check. In the 3D, step 5 lets you do the same with the L slider.
Key formulas and definitions
- F = −kx (k in N/m)
- T = 2π√(m/k), f = (1/2π)√(k/m)
- Series: 1/k = 1/k₁ + 1/k₂; Parallel: k = k₁ + k₂
- Vertical spring: k = mg/d, T = 2π√(d/g)
- Simple pendulum: T = 2π√(L/g)
Worked examples
1. A spring stretches by 5 cm when a 10 N force acts. Find k.
k = F/x = 10 / 0.05 = 200 N/m.
2. A 0.5 kg block is attached to a spring with k = 50 N/m. Find T and f.
T = 2π√(m/k) = 2π√(0.5/50) = 2π × 0.1 = 0.628 s. f = 1/T ≈ 1.59 Hz.
3. The mass on a spring is made 4 times. What happens to T?
T ∝ √m, so T becomes √4 = 2 times.
4. Two springs of 100 N/m each hold a 1 kg mass (a) in series, (b) in parallel. Find T in each case.
(a) k = 50 N/m, T = 2π√(1/50) ≈ 0.889 s. (b) k = 200 N/m, T = 2π√(1/200) ≈ 0.444 s.
5. A mass hung on a vertical spring stretches it by 2.5 cm. Find the period of vertical oscillations (g = 9.8 m/s²).
T = 2π√(d/g) = 2π√(0.025/9.8) = 2π × 0.0505 ≈ 0.317 s.
6. Find the time period of a simple pendulum of length 1 m (g = 9.8 m/s²).
T = 2π√(1/9.8) = 2π × 0.319 ≈ 2.01 s.
7. Find the length of a seconds pendulum (T = 2 s) where g = 9.8 m/s².
L = gT²/4π² = 9.8 × 4 / 39.48 ≈ 0.993 m.
8. A pendulum has T = 2 s on Earth. What is T on the Moon, where g is 1/6 of Earth’s?
T ∝ 1/√g, so T_moon = 2 × √6 ≈ 4.9 s.
Common mistakes
- Thinking a heavier bob swings faster. The mass cancels in a pendulum; only L and g matter.
- Using T = 2π√(L/g) for big swings. It is correct only for small angles (sinθ ≈ θ).
- Adding springs in series like resistors in series. For springs in series, add 1/k, not k.
- Measuring L only up to the hook. L is from the point of support to the centre of the bob.