CP5c – Vibrating String: Effect of Mass per Unit Length

Theory — Effect of Mass per Unit Length on Frequency

In this experiment you keep vibrating length L and tension T constant, and investigate how the resonant frequency f changes with mass per unit length μ by using strings of different types.

What You Are Investigating

The mass per unit length μ (also called linear density) is a property of the string itself — it depends on the material and the diameter. A thicker or denser string has a higher μ. Different strings are used for each reading while L and T remain fixed.

In CP5c you change the string type only. L and T stay fixed throughout.

The Key Relationship

From f = (1/2L)√(T/μ), when L and T are constant:

f ∝ 1/√μ i.e. f = k/√μ where k = √T/(2L) is a constant

Quadrupling μ halves the frequency. A string with 4× the mass per unit length vibrates at half the frequency under the same conditions.

Plotting f against 1/√μ gives a straight line through the origin. The gradient equals k = √T/(2L), which allows T to be calculated from a known L, or L from a known T.

Measuring μ

μ = m / ℓ (kg/m)

m = mass of the string sample (kg) · ℓ = length of the sample (m)
In practice: cut exactly 1.00 m of string, weigh it on an electronic balance. μ = mass in kg.

The simulation uses five string types with known μ values. In a real experiment you would measure μ for each string before starting.

Fixed Variables in This Experiment

Vibrating length L — clamp the string at the same position for every string type. Re-check L when you replace the string.
Tension T — use the same hanging masses for every string type. Do not add or remove masses between readings.

Procedure

Two methods to find resonance for each string type. Select the one your school uses.

Equipment

Signal generator · Vibration generator · At least 3 different strings (different materials or gauges) · Pulley and clamp stand · Slotted masses and hanger · Metre rule · Electronic balance (to measure μ)

Fixed throughout: Length L and tension T (same masses). Varied: String type (mass per unit length μ).

Method A — Fixed Frequency

Set the signal generator to a fixed frequency. Fix L and T. Change the string type. For each string, adjust L slightly until resonance appears at the fixed frequency, then record L and μ. This gives pairs of (f, L, μ) where f is fixed and L varies with μ.

1

Measure μ for each string before starting

Cut a 1.00 m sample of each string and weigh it on an electronic balance. Record the mass in grams — this equals μ in g/m. Do this for all strings before beginning the resonance measurements.

💡 Label each string sample clearly. Keep the offcuts so you can re-measure if needed.

2

Set up with the first string and fixed frequency

Install the thinnest string (lowest μ). Apply the fixed tension T (hanging masses). Set the signal generator to a chosen frequency (e.g. 200 Hz). Set L to a starting length. Switch the output ON.

💡 Start with the thinnest string — it has the highest resonant frequency, so you need a longer L to bring resonance to a convenient frequency.

3

Adjust L until resonance at the fixed frequency

Move the bridge to change L until a clear loop appears at the fixed frequency. Measure and record L at resonance. Record μ and f.

💡 Since f is fixed and μ changes between strings, L must also change to maintain resonance — this is the expected behaviour from f = (1/2L)√(T/μ).

4

Replace the string and repeat

Carefully remove the string and replace with the next type (higher μ). Re-apply the same tension T. Adjust L to find resonance at the same fixed frequency. Record the new L and μ.

💡 Thicker string = higher μ = lower wave speed = shorter resonant length at the same frequency.

5

Repeat for all string types

Work through all available strings. Collect at least 5 pairs of (μ, f). Calculate 1/√μ for each. In the simulation, the resonant frequency changes automatically when you select a different string — use Fixed Frequency mode.

💡 You need strings with a wide range of μ values for a reliable graph. Aim for the highest μ to be at least 4× the lowest μ.

6

Plot f against 1/√μ

Calculate 1/√μ for each reading (with μ in kg/m). Plot f (y-axis) against 1/√μ (x-axis). Gradient = √T/(2L).

💡 The units of 1/√μ are (kg/m)⁻½ = m½/kg½. The graph passes through the origin — confirming f ∝ 1/√μ.

Method B — Sweep Frequency (Recommended)

Fix L and T. Install each string in turn. Sweep the signal generator frequency to find the resonant frequency for that string. Each string gives one reading of (μ, f).

1

Measure μ and set fixed L and T

Measure μ for each string before starting (weigh 1.00 m samples). Set the vibrating length L (e.g. 0.60 m) and tension T (e.g. 300 g total). Write down L and T — they must not change.

💡 Fix T first by adding masses, then clamp the string at the correct L. Re-check both after every string replacement.

2

Install the thinnest string and sweep

Install the string with the smallest μ (e.g. thin nylon). Set up under the fixed L and T. Start at low frequency and sweep upward. Thin string = high wave speed = high resonant frequency — you may need to sweep past 400 Hz.

💡 Predict the approximate resonant frequency first: f ≈ (1/2L)√(T/μ). This tells you where to search.

3

Fine-tune and record

When a clear loop appears, use ±1 Hz nudge buttons to find the exact resonant frequency f₁. Record: string label, μ (g/m), f₁ (Hz), 1/√μ.

💡 The resonance indicator turns green when within ~1.5% of the true resonant frequency.

4

Replace with the next string and sweep again

Remove the string and replace with the next type (higher μ). Re-clamp at exactly the same L and re-apply the same T. Sweep frequency — the resonant frequency will be lower than before.

💡 In the simulation, click the next string button — L and T stay fixed. The resonance indicator shows how close the current frequency is to resonance.

5

Repeat for all 5 string types

Work through all strings from thinnest to thickest. As μ increases, f₁ decreases. Record μ and f₁ for each. You need at least 5 different string types for a reliable graph.

💡 The relationship is f ∝ 1/√μ. Quadrupling μ halves f. Check: thin nylon (0.50 g/m) to thick nylon (2.50 g/m) is a factor of √5 ≈ 2.24 reduction in f.

6

Plot f against 1/√μ

Calculate 1/√μ for each reading. Plot f (y-axis) against 1/√μ (x-axis). The gradient = √T/(2L). From the gradient and known T and L, verify the result.

💡 You can verify: gradient × 2L = √T. Calculate √T from your hanging masses and compare.

Fixed Variables

Vibrating length L0.60 m

Tension T2.94 N (300 g)

√T/(2L)—

Variable — String Type (μ)

Resonance Method

Harmonic

Resonance status

—

Resonant f

—

μ

—

1/√μ

—

Harmonic n

—

Readings recorded: 0 / 5 minimum

Amplitude response — ▲ marks current frequency relative to resonance

Results Data Table

Fixed: L = 0.60 m, T = 2.94 N. Varied: string type (μ). Collect one reading per string type.

#	String	n harmonic	μ / g m⁻¹	√μ / (g/m)½	1/√μ	f / Hz	f₁ = f/n / Hz
No data yet.

Graph & Analysis — f vs 1/√μ

Plot of resonant frequency f (y-axis) against 1/√μ (x-axis). A straight line through the origin confirms f ∝ 1/√μ. Gradient = √T/(2L).

√T/(2L) from gradient

—Hz·(g/m)½

Graph parameters

Gradient (= √T/2L)—

R²—

Expected √T/(2L)—

% difference—

Interpretation

Record at least 3 readings to see analysis.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

Explain why, in this experiment, you must replace the entire string rather than simply adding mass to it to change μ.

Adding mass to the string (for example by taping small weights along it) would change the total mass but also change the distribution of mass — the string would no longer have a uniform mass per unit length. The equation f = (1/2L)√(T/μ) assumes μ is constant and uniform along the entire length. A non-uniform string would not produce a clean stationary wave with clearly defined nodes and antinodes, and the resonant frequency would not follow the simple relationship. Therefore the only valid way to change μ is to replace the entire string with one of different linear density.

Q2

Show that the gradient of the f vs 1/√μ graph equals √T/(2L), and state what this allows you to determine experimentally.

From f = (1/2L)√(T/μ) = (√T/2L) × (1/√μ). This is of the form f = m × (1/√μ) where m = √T/(2L) is constant (since T and L are fixed). Therefore the gradient of f vs 1/√μ equals √T/(2L). Rearranging: T = (2L × gradient)². Since L is known, this allows the tension T to be calculated from the graph gradient alone — and compared with the value T = Mg from the hanging masses, providing an independent verification.

Q3

Predict the resonant frequency of a string with μ = 5.00 g/m, L = 0.60 m and T = 2.94 N. Show your working.

Using f = (1/2L)√(T/μ): μ = 5.00 g/m = 5.00 × 10⁻³ kg/m, L = 0.60 m, T = 2.94 N. Wave speed: v = √(T/μ) = √(2.94/5.00×10⁻³) = √588 = 24.2 m/s. Fundamental frequency: f₁ = v/(2L) = 24.2/(2×0.60) = 24.2/1.20 = 20.2 Hz. This is quite low — in practice you would need to use a longer string or higher tension to produce an easily visible fundamental mode at this μ.

Q4

Why does the graph of f vs 1/√μ have to pass through the origin? What would a positive y-intercept indicate?

When 1/√μ = 0, μ → ∞. An infinitely heavy string per unit length would have zero wave speed (v = √(T/μ) → 0 as μ → ∞), so f = v/2L → 0. Therefore the line must pass through the origin. A positive y-intercept would mean the string produces a non-zero frequency even when 1/√μ → 0, which is physically impossible. It could indicate a systematic error in measuring μ (e.g. using the wrong length for the sample), or that the string has significant stiffness, which adds a restoring force independent of tension.

Q5

Suggest two practical difficulties specific to this experiment that do not apply to CP5a or CP5b, and explain how each could be overcome.

1. Accurately measuring μ for each string — if the string sample is not exactly 1.00 m, or the balance has a zero error, μ will be incorrect, introducing a systematic error in the x-variable. Overcome by: cutting several 1.00 m lengths and weighing them all together, then dividing by the number of samples; using a four-decimal-place balance; and measuring the sample length with a steel rule rather than a tape measure. 2. Ensuring L and T are truly identical for each string replacement — when the string is removed and replaced, the clamp position can shift slightly, changing L, and the string may grip the pulley differently, altering the effective T. Overcome by: marking the clamp position precisely with a scribe mark, measuring L before and after each string replacement, and ensuring the string lies in the same groove on the pulley each time.