Determining an unknown mass from resonance

Practical method

Aim

Find the resonant frequency of a driven mass–spring system for known masses and use a calibration graph to find an unknown mass.

Variables

Independent variable: mass on the spring
Dependent variable: resonant frequency f₀
Control variables: the same spring; the same driving amplitude.

Apparatus

• Spring with mass hanger
• Vibration generator driven by a signal generator
• Metre rule behind the mass
• Slotted masses and an unknown mass

Method

1. Hang the spring from the vibration generator with a 0.100 kg mass.
2. Slowly increase the driving frequency, watching the amplitude of the mass.
3. Record the frequency at which the amplitude is largest: the resonant frequency. Approach from above and below and average.
4. Repeat for masses up to 0.500 kg.
5. Plot \(\dfrac{1}{f_{0}^{2}}\) against m (a straight line, since f₀ = \(\dfrac{1}{2\pi }\)\(\sqrt{\dfrac{k}{m}}\)); find the unknown mass from its resonant frequency.

Sample results

Spring constant about 40 N m⁻¹. Unknown mass resonates at 2.00 Hz.

Analysis

\(\dfrac{1}{f_{0}^{2}}\) is proportional to m with gradient 0.987 s² kg⁻¹ (= \(\dfrac{4\pi ^{2}}{k}\), so k = 40 N m⁻¹). The unknown mass has \(\dfrac{1}{f_{0}^{2}}\) = 0.250 s², so m = \(\dfrac{0.250}{0.987}\) = 0.253 kg.

Uncertainty

The resonance peak is broad, so f₀ is judged to about ±0.05 Hz (2–3%). Approaching from both sides and averaging reduces the bias; \(\dfrac{1}{f_{0}^{2}}\) doubles the percentage uncertainty.

Errors and improvements

Risk assessment

Common mistakes

• Reading the frequency where the motion first becomes visible.
• Plotting f₀ against m and expecting a line.
• Not repeating from above and below.

Worked analysis

Unknown: f₀ = 2.00 Hz, so \(\dfrac{1}{f_{0}^{2}}\) = 0.250 s². From the gradient 0.987 s² kg⁻¹, m = \(\dfrac{0.250}{0.987}\) = 0.253 kg.

Common misconception

At resonance the driving frequency equals the natural frequency; the amplitude is largest but the driver's amplitude is unchanged.