Core practical 16: finding a mass from resonant frequencies

Method

• Clamp a vibration generator, driven by a signal generator, above the bench. Hang the spring from the vibration generator and hang the mass hanger from the lower end of the spring.
• For each known mass, slowly increase the frequency and find the frequency f at which the amplitude is greatest (resonance). Use small frequency steps near the peak.
• Repeat for at least five masses, and for the unknown mass.
• Alternatively, time 10 free oscillations to find T and f = 1 / T: at resonance, the driving frequency equals the natural frequency.

Analysis

f = (1/2π) sqrt(k / m), so 1 / f² = 4π^2m / k (equivalently T² = 4π^2m / k). A graph of 1/f² (or T²) against m is a straight line through the origin with gradient 4π² / k. Read off the unknown mass from its value of 1/f², or calculate it using k from the gradient.

Errors

• The spring's own mass means the line may not pass exactly through the origin; using the graph rather than a single calculation deals with this.
• The resonance peak is easier to locate precisely with light damping and fine frequency control.

Worked example

A spring with k = 40 N/m resonates at 2.5 Hz with an unknown mass. Find the mass.

1. f = (1/2π) sqrt(k / m), so m = k / (4π^2f2)
2. m = 40 / (39.48 x 6.25)

Answer: m = 0.162 kg

Check yourself

1. A spring with k = 40 N/m resonates at 2.5 Hz with an unknown mass. Find the mass in kg, to 3 s.f.
   Show answer

   0.162 kg. m = k / (4π^2f2) = 40 / (39.48 x 6.25) = 0.162 kg.

2. A graph of 1/f² against m has a gradient of 0.95 s²/kg. Find the spring constant in N/m, to 3 s.f.
   Show answer

   41.6 N/m. Gradient = 4π² / k, so k = 39.48 / 0.95 = 41.6 N/m.

3. Using the same line through the origin (gradient 0.95 s²/kg), an unknown mass gives 1/f² = 0.19 s². Find the mass in kg.
   Show answer

   0.2 kg. m = 0.19 / 0.95 = 0.20 kg.

4. A 0.30 kg mass on a spring makes 10 oscillations in 7.0 s. Find the spring constant in N/m, to 3 s.f.
   Show answer

   24.2 N/m. T = 0.70 s; k = 4π^2m / T² = 39.48 x 0.30 / 0.49 = 24.2 N/m.

5. Which graph should give a straight line through the origin in this practical?
   • f against m
   • 1/f² against m
   • f² against m
   • f against 1/m
   Show answer

   1/f² against m. 1/f² = (4π²/k)m.

6. Why should the frequency be changed in small steps near resonance?
   • To stop the spring breaking
   • To keep the mass constant
   • To reduce the spring constant
   • The amplitude peak is narrow, so small steps locate it precisely
   Show answer

   The amplitude peak is narrow, so small steps locate it precisely. Large steps could miss the true peak.

7. The line on the graph does not quite pass through the origin. What is the most likely reason?
   • The signal generator is faulty
   • The spring has mass that also oscillates
   • The spring constant is zero
   • Resonance cannot occur with small masses
   Show answer

   The spring has mass that also oscillates. Part of the spring's mass adds to the oscillating mass.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Cognito: Edexcel International A level Physics specification and core practicals

Last updated 7 November 2026. Spotted a mistake? Tell us.