CP16 – Resonant Frequencies & Unknown Mass

Theory — Deriving the Period Equation

You should derive this expression yourself before beginning the practical.

Research Questions

What does Hooke's Law state? How is the spring constant k defined?
When a mass on a spring is displaced and released, what type of motion results? What is the restoring force?
Derive an expression for the period T of oscillation in terms of mass m and spring constant k.
How will you linearise your data to plot a straight-line graph? What will the gradient and intercept be?

Hooke's Law

F = kx

F = restoring force (N) · k = spring constant (N m⁻¹) · x = displacement from equilibrium (m)

When a mass hangs at rest on a spring, the spring extends by x₀ = mg/k. Displacing it further by x and releasing creates a restoring force F = −kx (where x is now measured from equilibrium) — simple harmonic motion.

Deriving T in Terms of m and k

For SHM, Newton's second law gives F = ma = −kx, so a = −(k/m)x. This is SHM with angular frequency ω = √(k/m). The period is:

T = 2π / ω = 2π √(m/k) Squaring: T² = (4π²/k) × m

This is of the form y = mx where y = T², x = m, and gradient = 4π²/k.

A graph of T² against m gives a straight line through the origin. The gradient = 4π²/k, so k = 4π²/gradient. Then the unknown mass m_x = T_x² × k / 4π² = T_x² / gradient.

Finding the Unknown Mass

Method 1 — from the graph: measure T for the unknown mass, calculate T², then read off m from the x-axis of the calibration graph (or calculate m = T²/gradient).

Method 2 — from k: calculate k = 4π²/gradient from the calibration line. Then use m_x = kT_x²/4π² with the measured T_x.

m_unknown = T²_unknown × k / (4π²) = T²_unknown / gradient

Measuring Period Accurately

Time 10 complete oscillations and divide by 10 — this reduces the percentage error in T by a factor of 10 compared to timing a single oscillation. Repeat three times and take the mean. The percentage uncertainty in T² = 2 × percentage uncertainty in T.

Procedure

Plan and carry out your investigation to find the unknown mass.

Equipment

Spiral spring · Clamp stand, boss and clamp · Slotted masses (100g, 200g, 300g, 400g, 500g, 600g) · Mass hanger · Unknown mass (labelled X) · Electronic balance · Stopwatch · Metre rule · Safety mat below masses

1

Write your plan first

Before starting: write down the equation relating T and m, state which graph you will plot to get a straight line, state how you will find m_unknown from the graph. The approximate unknown mass is ~300 g.

💡 Planning the graph before collecting data helps you choose an appropriate range of known masses. Use masses from ~100g to ~600g to bracket the unknown mass.

2

Measure the spring constant k (optional)

Hang masses on the spring and measure the extension for each. Plot F vs x — gradient = k. This gives an independent check on k from your T² vs m graph.

💡 The two values of k should agree within experimental uncertainty. A large discrepancy suggests the spring is not behaving as an ideal Hooke's Law spring (e.g. coils touching at low extensions, or exceeding the elastic limit at high loads).

3

Hang first known mass and displace gently

Hang 100 g on the spring. Pull down by about 1–2 cm and release. This is the amplitude — keep it small and consistent for all masses (large amplitudes can cause the mass to sway sideways or bounce non-linearly).

💡 Pull straight down — any sideways component causes pendulum motion mixed with spring oscillation. Release cleanly without giving an extra push.

4

Time 10 complete oscillations

Count oscillations from the equilibrium position passing upward. Start the stopwatch as the mass passes equilibrium going up (count 0), stop when it passes equilibrium going up for the 10th time. Divide by 10 to get T. Repeat 3 times, take the mean T. Calculate T².

💡 Starting and stopping the timer at the equilibrium position (fastest moving point) is easier and more accurate than timing from the extreme positions (where the mass is momentarily stationary — harder to judge exactly).

5

Repeat for all masses including Unknown X

Repeat for 200g, 300g, 400g, 500g, 600g, and Unknown X. Record T² for each. Also time 10 oscillations for the unknown mass at least 3 times.

💡 Keep the same small amplitude throughout. If the amplitude changes between runs, the period changes (for a spring obeying Hooke's Law it should not, but non-ideal springs do show small amplitude dependence).

6

Plot T² vs m and determine unknown mass

Plot T² (y-axis) against m in kg (x-axis). Draw best-fit line through origin. Read off m_unknown from the x-axis using T²_unknown. Compare with the balance measurement of the unknown.

💡 The line should pass through the origin — T = 0 when m = 0. A positive y-intercept suggests the spring itself has significant mass (which contributes to the effective oscillating mass). If this occurs, report it and discuss its significance.

      🌀 Select a known mass, set the amplitude, press ▶ Release to start oscillating, then time 10 oscillations and record.
    

Known Masses

Unknown Mass

Amplitude

Displacement A 2.0 cm

0.5 cm5 cm

Spring Properties

Oscillation Timer

—

Period T / s

Oscillations counted0

Elapsed time0.000 s

T² = T×T—

Readings recorded: 0/6 minimum

Results Data Table

Plot T² (y-axis) against m in kg (x-axis). Straight line through origin — gradient = 4π²/k.

#	Mass / g	Mass / kg	10T / s	T = 10T/10 / s	T² / s²	Note
No data yet — use the Simulation tab.

Graph & Analysis — T² vs m

Straight line through origin confirms T² = (4π²/k) × m. The unknown mass is read from the x-axis using its measured T².

Unknown mass determination

—g — from T² vs m graph

Graph parameters

Gradient (= 4π²/k)—

y-intercept (→ 0)—

R²—

k from gradient—

True k (actual)—

% error in k—

Interpretation

Record known mass readings then measure unknown T² to see analysis.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

Show that T² = (4π²/k)m follows from Newton's second law applied to a mass on a spring. State what graph you would plot to confirm this relationship and how the spring constant k is found from it.

When a mass m is displaced x from equilibrium on a spring with constant k, the restoring force is F = −kx. By Newton's second law: F = ma, so ma = −kx, giving a = −(k/m)x. This is SHM with ω² = k/m. The period T = 2π/ω = 2π√(m/k). Squaring: T² = 4π²m/k = (4π²/k)m. This is of the form y = mx (straight line through origin) where y = T², x = m, gradient = 4π²/k. Plot T² (y-axis) against m (x-axis) — the gradient gives k = 4π²/gradient. This is confirmed if the line passes through the origin (y-intercept = 0).

Q2

Explain why timing 10 oscillations and dividing by 10 gives a more accurate value of T than timing a single oscillation.

The main source of uncertainty in timing is human reaction time, typically ±0.1–0.2 s for starting and stopping a stopwatch. For a single oscillation of period T ≈ 0.8 s, the timing uncertainty is approximately ±0.2 s total (start and stop), giving a percentage uncertainty of 0.2/0.8 × 100 ≈ 25%. For 10 oscillations (10T ≈ 8 s), the same timing uncertainty ±0.2 s gives a percentage uncertainty of 0.2/8 × 100 ≈ 2.5%. Dividing by 10 gives T with 2.5% uncertainty — 10 times better. Note: the timing uncertainty in 10T is the same as for 1T (reaction time does not add up), but dividing by 10 reduces the absolute uncertainty in T by a factor of 10.

Q3

A student's T² vs m graph has a positive y-intercept rather than passing through the origin. Suggest a reason and explain how it could be corrected.

A positive y-intercept occurs when T² > 0 as m → 0. From T = 2π√(m_eff/k), if the spring itself has mass m_s, the effective oscillating mass is m_eff = m_hung + m_s/3 (where m_s/3 is the effective mass of the spring — from the theory of springs). So T² = (4π²/k)(m + m_s/3). This gives y-intercept = 4π²m_s/(3k) ≠ 0. The y-intercept gives m_s/3 = y-intercept × k/4π² from which the spring's mass can be estimated. The corrected approach is to use the x-intercept of the line (where T = 0) — this gives −m_s/3, confirming the spring mass contribution. The unknown mass is still correctly found from the gradient regardless of the intercept.

Q4

The student measures the spring constant by hanging masses and recording extensions, getting k = 18.2 N/m. From their T² vs m graph, they get k = 19.5 N/m. Calculate the percentage difference and suggest reasons for the discrepancy.

% difference = |19.5 − 18.2| / 18.2 × 100 = 1.3/18.2 × 100 = 7.1%. Possible reasons: 1. The spring mass contributes to the oscillating mass but not to the static extension, so the dynamic method (T² vs m) gives an effectively larger apparent k. 2. Measurement uncertainty in extension readings for the static method — extension must be read very carefully, and any parallax error adds systematic uncertainty. 3. The spring may not perfectly obey Hooke's Law — at low loads the coils may not be fully open, giving a lower k than at higher loads. 4. Damping — if the oscillation damps significantly, the observed period is slightly longer than the true SHM period (T_damped = T₀/√(1 − ζ²) where ζ is the damping ratio), making k appear smaller from the dynamic method.

Q5

Explain why the graph of T² against m must pass through the origin, and what it would mean physically if the line had a negative x-intercept.

From T² = (4π²/k)m: when m = 0 there is no oscillating mass and T = 0, so T² = 0 and the line passes through the origin. A negative x-intercept means that T² = 0 at a negative value of m — physically meaningless unless interpreted as the effective spring mass. If the best-fit line has equation T² = gradient × (m + m_eff), then the x-intercept is at m = −m_eff. This means the spring itself contributes positively to the oscillating mass (m_eff = m_s/3 for a uniform spring), shifting the line so it crosses the x-axis at a negative value. In practice this means: even a massless spring (m_hung = 0) would oscillate with a non-zero period due to the spring's own mass — but since you can't actually have a massless hung mass in practice, you can only infer this from extrapolating the fitted line.