PhysicsISC Class 11 · Oscillations

Spring-Mass System and Simple Pendulum: Periods and Effective g

Updated 17 October 2026 · Fact-checked

A mass m on a spring of force constant k oscillates with T = 2π√(m/k), horizontally or vertically. A simple pendulum of length l swinging through small angles has T = 2π√(l/g). Both periods are independent of amplitude; the pendulum period depends on the effective g, not on the mass.

Animation of a mass hanging on a spring, beside a phase-space circle labelled Orbit with position and velocity axes. — Watch the mass on the spring oscillate up and down while its position and velocity trace a closed orbit in phase space.
Picture: Mazemaster · Public domain · Wikimedia Commons

Notice that T depends on m and k for the spring, but on l and g for the pendulum, and neither depends on amplitude.

Illustration drawn for QuestionBuddy.

Understand Spring-Mass System and Simple Pendulum

Spring-mass system. For a mass m attached to a spring with restoring force F = −kx, a = −(k/m)x, so ω = √(k/m) and T = 2π√(m/k). On a smooth horizontal surface the equilibrium is at the natural length. For a vertical spring, the equilibrium is where mg = kx₀, and the motion about that point has the same T = 2π√(m/k) = 2π√(x₀/g), where x₀ is the static extension. Gravity only shifts the equilibrium.

Combinations of springs. In parallel the force constants add: k = k₁ + k₂. In series, 1/k = 1/k₁ + 1/k₂. A spring of force constant k cut into two equal halves gives each half a force constant 2k, because each half stretches half as much for the same force.

Simple pendulum. A point mass on a light inextensible string of length l, oscillating through a small angle θ (θ in radians, so that sinθ ≈ θ). The tangential component of weight gives a = −(g/l)s, a = −ω²s with ω² = g/l, so T = 2π√(l/g). The period is independent of the mass of the bob and of the amplitude (for small angles); the length is measured from the point of suspension to the centre of the bob.

Seconds pendulum. T = 2 s, so l = gT²/4π² ≈ 0.99 m at g = 9.8 m s⁻².

Effective g. If the point of suspension accelerates, replace g by g_eff: upward acceleration a gives g_eff = g + a, downward a gives g − a, and free fall gives g_eff = 0 (no oscillation). A pendulum clock loses time if its period increases.

Energy. The energy of the spring-mass system is ½kA², as for any SHM, and the speed at the mean position is A√(k/m).

Spring-Mass System and Simple Pendulum: formulas and key facts

Spring-mass: $T = 2 π \sqrt (m / k) ∣ T = 2 π \sqrt (x_{0} / g)$; x₀ = static extension (vertical).
Springs: $P a r a l l e l : k = k_{1} + k_{2} ∣ S er i es : 1/ k = 1/ k_{1} + 1/ k_{2}$; Half spring: k′ = 2k.
Simple pendulum: $T = 2 π \sqrt (l / g)$; Small angles; l to the centre of the bob.
Effective g: $g_{e} f f = g \pm a$; Plus for upward acceleration of the support.

Spring-Mass System and Simple Pendulum in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A shop-floor weighing spring

Suppose a 0.4 kg mass hung on a vertical spring stretches it by 4.9 cm at rest (g = 9.8 m s⁻²).

Find the force constant and the period of vertical oscillations.

Try it first, then show the solutionSolution

k = mg/x₀ = 0.4 × 9.8/0.049 = 80 N m⁻¹.
T = 2π√(m/k) = 2π√(0.4/80) = 2π × 0.0707 = 0.444 s.
Check: 2π√(x₀/g) = 2π√(0.005) = 0.444 s.

Answer: k = 80 N m⁻¹; T = 0.444 s.

Memory hook: Static extension alone gives the period: T = 2π√(x₀/g).

Tricks to remember Spring-Mass System and Simple Pendulum

Short memory aids. Say each one aloud once, then use it on a question the same day.

Stiffer, faster
T ∝ √(m/k): more mass slows the oscillation, a stiffer spring speeds it up.
Use it when: Comparing springs.
g_eff for lifts
Upward acceleration makes the pendulum quicker; downward acceleration makes it slower.
Use it when: Pendulum in a lift.

Spring-Mass System and Simple Pendulum: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Independence of g · 1 mark · MCQBoard level

The time period of a mass-spring system on a vertical spring on the Earth and on the Moon is:
(a) larger on the Moon
(b) smaller on the Moon
(c) the same
(d) zero on the Moon

Try it first, then show the solutionSolution

T = 2π√(m/k) does not contain g.

Answer: (c) the same

Example 2: Spring-mass period · 3 marks · Short answerBoard level

A 0.5 kg mass is attached to a spring of force constant 200 N m⁻¹ on a smooth horizontal table. Find the period and frequency of its oscillation.

Try it first, then show the solutionSolution

T = 2π√(m/k). (1 mark)
= 2π√(0.5/200) = 2π × 0.05 = 0.314 s. (1 mark)
f = 1/T = 3.18 Hz. (1 mark)

Answer: T = 0.314 s; f = 3.18 Hz.

Example 3: Vertical spring · 4 marks · Short answerHarder board level

A 0.4 kg mass hung from a vertical spring stretches it by 4.9 cm. It is then pulled down a further 3 cm and released (g = 9.8 m s⁻²). Find the force constant, the period, the maximum speed and the maximum force in the spring.

Try it first, then show the solutionSolution

k = mg/x₀ = 3.92/0.049 = 80 N m⁻¹. (1 mark)
T = 2π√(m/k) = 2π√(0.005) = 0.444 s. (1 mark)
v_max = Aω = 0.03 × √(80/0.4) = 0.03 × 14.14 = 0.424 m s⁻¹. (1 mark)
F_max = k(x₀ + A) = 80 × 0.079 = 6.32 N. (1 mark)

Answer: k = 80 N m⁻¹; T = 0.444 s; v_max = 0.42 m s⁻¹; F_max = 6.3 N.

Example 4: Springs in series and parallel · 4 marks · Short answerHarder board level

Two springs of force constants 100 N m⁻¹ and 300 N m⁻¹ support a 1 kg mass. Find the period when they are joined (i) in series and (ii) in parallel.

Try it first, then show the solutionSolution

(i) 1/k = 1/100 + 1/300 = 4/300, so k = 75 N m⁻¹. (1 mark)
T = 2π√(1/75) = 0.7255 s. (1 mark)
(ii) k = 100 + 300 = 400 N m⁻¹. (1 mark)
T = 2π√(1/400) = 2π/20 = 0.314 s. (1 mark)

Answer: Series: 0.726 s; parallel: 0.314 s.

Example 5: Pendulum clock · 5 marks · Long answerHOTS / competency

A simple pendulum 1.00 m long (g = 9.8 m s⁻²) is the timekeeper of a clock. Find its period. If its length increases by 1%, by how much per cent and per day does the clock lose? If instead g is reduced by 0.2% on a hill, what happens in a day?

Try it first, then show the solutionSolution

T = 2π√(1/9.8) = 2.007 s. (1 mark)
l → 1.01l gives T → √1.01 T = 1.00499T, an increase of 0.50%. (1 mark)
The clock loses 0.50% of 86 400 s = 431 s ≈ 7.2 minutes a day. (1 mark)
g reduced by 0.2%: T ∝ 1/√g, so T increases by 0.1%. (1 mark)
Loss = 0.1% of 86 400 = 86 s per day. (1 mark)

Answer: T = 2.007 s; loses about 431 s per day for the longer pendulum and about 86 s per day at the lower g.

Where students lose marks in Spring-Mass System and Simple Pendulum

The slips examiners see most often, and what to write instead.

Mistake: Using the amplitude in T
Writing T = 2π√(l/g) with a dependence on the swing angle.
Do this: The period of a small-angle pendulum does not depend on the amplitude or mass.
Usually costs: 1 mark.
Mistake: Wrong combination of springs
Adding force constants for springs in series.
Do this: Parallel: add k. Series: add 1/k.
Usually costs: 2 marks.
Mistake: Length of the pendulum
Measuring l to the bottom of the bob.
Do this: l is from the suspension point to the centre of the bob.
Usually costs: 1 mark.

Exam tips

State the small-angle condition when giving T = 2π√(l/g).
Use T = 2π√(x₀/g) as a quick check for vertical springs.
Write g_eff explicitly in lift problems.
Give T in seconds and f in hertz.

Practice questions on Spring-Mass System and Simple Pendulum

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The length of a simple pendulum is measured from the point of suspension to:
(a) the lowest point of the bob
(b) the centre of the bob
(c) the top of the bob
(d) the middle of the string
Hint
The bob is treated as a point mass at its centre.
Answer
(b) the centre of the bob
Q2 · 2 marks · Short answerBoard level
Find the length of a seconds pendulum at a place where g = 9.8 m s⁻².
Hint
Seconds pendulum: T = 2 s.
Answer
l = gT²/4π² = 9.8 × 4/39.48 = 0.993 m.
Q3 · 3 marks · Short answerHarder board level
A mass m on a spring has a period of 1 s. The spring is cut into two equal halves and the same mass is attached to one half. Find the new period.
Hint
Half the spring has twice the force constant.
Answer
k′ = 2k, so T′ = T/√2 = 0.707 s.
Q4 · 3 marks · Short answerHarder board level
A simple pendulum 2.0 m long swings on the Moon (g = 1.6 m s⁻²) and on the Earth (g = 9.8 m s⁻²). Find its period in each place.
Hint
T = 2π√(l/g).
Answer
Moon: T = 2π√(2/1.6) = 7.02 s. Earth: T = 2π√(2/9.8) = 2.84 s.
Q5 · 3 marks · Short answerHarder board level
A 2 kg mass on a smooth horizontal spring (k = 50 N m⁻¹) oscillates with an amplitude of 0.1 m. Find the total energy, the maximum speed and the period.
Hint
E = ½kA².
Answer
E = ½ × 50 × 0.01 = 0.25 J. v_max = A√(k/m) = 0.1 × 5 = 0.5 m s⁻¹. T = 2π√(2/50) = 1.26 s.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a simple pendulum of length 1.0 m hangs inside a lift (g = 9.8 m s⁻²).
(i) Find its period when the lift is at rest. (1 mark)
(ii) Find its period when the lift accelerates upward at 2 m s⁻². (1 mark)
(iii) Find its period when the lift accelerates downward at 2 m s⁻². (1 mark)
(iv) What happens if the lift falls freely? (1 mark)
Hint
Use g_eff = g + a for upward and g − a for downward acceleration.
Answer
(i) T = 2π√(1/9.8) = 2.01 s. (ii) g_eff = 11.8: T = 1.83 s. (iii) g_eff = 7.8: T = 2.25 s. (iv) g_eff = 0, so the bob does not oscillate (T → ∞).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Time period of a mass on a spring?Show answer
T = 2π√(m/k).
2.Time period of a simple pendulum?Show answer
T = 2π√(l/g).
3.Equivalent k for springs in parallel?Show answer
k₁ + k₂.
4.Does T of a pendulum depend on the mass?Show answer
No.

Spring-Mass System and Simple Pendulum: frequently asked questions

Why is the pendulum only SHM for small angles?

The restoring force is mg sinθ, which is proportional to θ only when sinθ ≈ θ (θ small, in radians).

Does a vertical spring oscillate with a different period from a horizontal one?

No. Gravity only shifts the equilibrium position; T = 2π√(m/k) in both.

What does the pendulum do in a satellite?

In orbit, everything is in free fall, so g_eff = 0 and the pendulum does not oscillate.