CP5b – Vibrating String: Effect of Tension

Theory — Effect of Tension on Frequency

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

What You Are Investigating

The tension T in the string is controlled by the weight of hanging masses: T = Mg. Increasing the tension increases the wave speed on the string, which increases the resonant frequency at a fixed length.

In CP5b you change T only by adding or removing 100 g slotted masses. L and μ stay fixed throughout.

The Key Relationship

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

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

Quadrupling the tension doubles the frequency. Halving the tension reduces the frequency by a factor of √2 ≈ 1.41.

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

Tension from Hanging Masses

T = Mg

M = total hanging mass in kg (hanger + discs) · g = 9.81 m s⁻²
Each 100 g disc adds T = 0.100 × 9.81 = 0.981 N

Fixed Variables in This Experiment

Vibrating length L — keep the clamp or bridge at the same position for every reading. Do not adjust L between readings.
Mass per unit length μ — use the same string throughout. Do not swap strings.

Procedure

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

Equipment

Signal generator · Vibration generator · String · Pulley and clamp stand · Slotted masses (100 g) and hanger · Metre rule

Fixed throughout: Vibrating length L and string type (μ). Varied: Tension T (number of slotted masses).

Method A — Fixed Frequency

Set the signal generator to a fixed frequency. Add or remove slotted masses to change T until resonance appears at that frequency. Repeat at several different frequencies.

1

Set up the apparatus with fixed L

Clamp the string at a fixed vibrating length L (e.g. 0.60 m). Measure and record L. This must not change. Start with the hanger only (100 g) as minimum tension.

💡 Mark the bridge/clamp position on the bench so L cannot shift accidentally.

2

Set a fixed frequency

Choose a frequency on the signal generator (e.g. 100 Hz). Write it down. Switch the output ON. This frequency stays constant while you search for the resonant tension.

💡 Choose a frequency that gives resonance somewhere in the range 100–700 g total mass. You can estimate this from f = (1/2L)√(T/μ).

3

Add masses until resonance appears

Add 100 g discs one at a time. After each addition, wait a moment for the string to settle. Resonance appears as a sudden increase in amplitude with a clear loop pattern visible.

💡 Resonance may not occur at exact 100 g intervals with a fixed frequency. If it falls between two values, interpolate or use a slightly different frequency.

4

Record T and f

Note the total hanging mass M and calculate T = Mg. Record T, √T and the fixed frequency f.

💡 Sweep method (Method B) is much easier for investigating T — it avoids needing resonance to fall at an exact mass value.

5

Change frequency and find new resonant T

Change to a new fixed frequency. Adjust the number of masses until resonance appears again. Repeat for at least 6 different tensions across a wide range.

💡 Vary T over as wide a range as possible — e.g. 100 g to 700 g total mass — to get a clear gradient on the f vs √T graph.

6

Plot f against √T

Calculate √T for each reading. Plot f (y-axis) against √T (x-axis). The gradient = 1/(2L√μ).

💡 The line must pass through the origin — confirming f ∝ √T.

Method B — Sweep Frequency (Recommended)

Fix the vibrating length L and add a set number of masses. Sweep the frequency upward until resonance appears. Record f for each tension. This is the preferred method for investigating T.

1

Fix L and set minimum tension

Clamp the string at a fixed length L (e.g. 0.60 m). Start with the hanger only (100 g). L stays constant for every reading in this experiment.

💡 Re-check L before every reading — the string can creep slightly as tension changes.

2

Sweep from low frequency to find f₁

Start at 20 Hz and increase slowly. At the resonant frequency a clear single loop appears with maximum amplitude. Fine-tune with ±1 Hz nudge buttons to find the exact resonant frequency f₁.

💡 With low tension the resonant frequency is low — search below 100 Hz first with a light hanger.

3

Record T and f₁

Record: total mass M (kg), tension T = Mg (N), √T, and resonant frequency f₁ (Hz). This is your first reading.

💡 Each 100 g disc adds T = 0.981 N. The new resonant frequency ≈ old f × √(new T / old T) — use this to predict where to search next.

4

Add 100 g and sweep again

Add one 100 g disc. The resonant frequency increases because f ∝ √T. Sweep from a frequency slightly below the predicted new f₁ to find it quickly.

💡 Adding one disc increases T by about 18% (100 g on 100 g hanger → 200 g total) which increases f by √(200/100) = √2 ≈ 1.41 times.

5

Repeat for at least 6 tensions

Add discs progressively (100 g, 200 g, 300 g … up to 700 g or more total). Find f₁ for each. A wide range of T gives a more reliable gradient on the graph.

💡 Aim for a factor of at least 4–5 difference between minimum and maximum T for a clear graph.

6

Plot f against √T

Calculate √T for each reading. Plot f (y-axis) against √T (x-axis). The best-fit line through the origin has gradient = 1/(2L√μ). From the gradient and known L, calculate μ.

💡 Verify your result: calculate μ = 1/(2L × gradient)² and compare with the string's known μ (weigh a 1 m length on a balance).

Fixed Variables

String typeMedium nylon

μ1.20 g/m

Vibrating length L0.60 m

Variable — Tension T (slotted masses)

Hanging masses 200 g

        T = Mg = — N  ·  √T = — N½
      

Resonance Method

Harmonic

Resonance status

—

Resonant f

—

Tension T

—

√T

—

Harmonic n

—

Readings recorded: 0 / 6 minimum

Amplitude response — ▲ marks current frequency relative to resonance

Results Data Table

Fixed: L = 0.60 m, μ = 1.20 g/m. Varied: tension T (hanging masses).

#	n harmonic	Mass M / g	T = Mg / N	√T / N½	f / Hz	f₁ = f/n / Hz
No data yet.

Graph & Analysis — f vs √T

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

Calculated μ from gradient

—g/m from gradient

Graph parameters

Gradient (= 1/(2L√μ))—

R²—

Expected μ—

% difference—

Interpretation

Record at least 3 readings to see analysis.

Discussion Questions

Write your answers and reveal model answers when ready.

Q1

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

From f = (1/2L)√(T/μ) = (1/(2L√μ)) × √T. This is of the form f = m√T where m = 1/(2L√μ) is the gradient. Since L is known (fixed and measured), rearranging gives μ = 1/(2L × gradient)². This allows the mass per unit length of the string to be determined experimentally from the graph gradient, which can then be compared with the value measured directly by weighing a known length of string.

Q2

A student doubles the tension in the string. By what factor does the resonant frequency change? Show your working.

From f ∝ √T: if T doubles, f changes by a factor of √2 ≈ 1.41. So the new frequency is 1.41 times the original frequency. Example: if f₁ = 150 Hz at T = 2.0 N, then at T = 4.0 N: f₂ = 150 × √(4.0/2.0) = 150 × √2 = 150 × 1.414 = 212 Hz. Note: you need to quadruple the tension (multiply T by 4) to double the frequency.

Q3

Explain why the sweep frequency method is recommended over the fixed frequency method when investigating tension T.

In the fixed frequency method, you must adjust the hanging mass until resonance happens to occur at the fixed frequency. This is impractical because resonance may not fall at a convenient mass value (e.g. an exact multiple of 100 g), requiring fine adjustment of mass which is difficult with slotted discs. In contrast, the sweep method fixes the mass (and therefore T precisely) and sweeps frequency to find f — you can read f directly from the signal generator display to high precision. This gives smaller uncertainty in the x-variable (√T is exact) and smaller uncertainty in f than in mass interpolation.

Q4

Explain why the f vs √T graph must pass through the origin, and what it would mean if it had a positive y-intercept.

From f = k√T, when T = 0 the wave speed v = √(T/μ) = 0, so f = v/2L = 0. Therefore f = 0 when √T = 0, and the line must pass through the origin. A positive y-intercept would mean the string resonates at some frequency even with zero tension, which is physically impossible. It would indicate a systematic error, most likely that the string has some stiffness (it behaves partly like a rod rather than an ideal flexible string), which adds an extra restoring force at zero tension. This effect is more pronounced with stiffer strings such as thick steel wire.

Q5

State two sources of error specific to this experiment and explain how each could be reduced.

1. String stretching — as tension increases, the string stretches slightly, increasing L and changing μ. Both effects alter the resonant frequency and introduce a systematic error. Reduce by re-measuring L before each reading and using a stiffer string with lower elasticity (e.g. steel wire for the higher-tension readings). 2. String creep at the clamp — the string may slip slightly at the clamp or over the pulley as tension changes, altering the effective vibrating length. Reduce by using a firm clamp with a groove or saddle, and by checking that the string does not slip during the measurement by re-measuring L after adding each mass.