CP5a

Vibrating String — Effect of Length on Resonant Frequency

Edexcel IAL · AS Physics

Theory — Effect of Length on Frequency

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

What You Are Investigating

A string under tension, fixed at both ends, vibrates in a stationary wave pattern when driven at a resonant frequency. The simplest pattern (fundamental mode) has one loop. The resonant frequency depends on three things: the length L, the tension T, and the mass per unit length μ of the string.

In CP5a you change L only. T and μ stay fixed throughout.

The Key Relationship

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

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

Doubling the length halves the frequency. Halving the length doubles the frequency.

Plotting f against 1/L gives a straight line through the origin. The gradient equals k = √(T/μ)/2, which equals half the wave speed on the string.

Stationary Waves and Harmonics

At the fundamental frequency (1st harmonic) there is one loop and λ = 2L. Higher harmonics occur at integer multiples of the fundamental:

1st harmonic: f₁ = v / 2L (1 loop, λ = 2L) 2nd harmonic: f₂ = v / L (2 loops, λ = L) 3rd harmonic: f₃ = 3v / 2L (3 loops, λ = 2L/3)

v = √(T/μ) = wave speed on the string · All harmonics satisfy the same f ∝ 1/L relationship.

The nodes (zero displacement) are at the fixed ends and at evenly spaced points along the string. The antinodes (maximum displacement) are midway between nodes.

Fixed Variables in This Experiment

Tension T — set by the hanging mass. Keep the same masses throughout every reading.
Mass per unit length μ — determined by the string material and diameter. Use the same string throughout. Do not swap strings mid-experiment.

Procedure

Two methods to find resonance. Select the one your school uses — or try both.

Equipment

Signal generator · Vibration generator (mechanical oscillator) · String · Pulley and clamp stand · Slotted masses and hanger · Metre rule · String of known μ (weigh a 1.00 m length and divide by length)

Fixed throughout: Tension T (same masses on hanger) and mass per unit length μ (same string). Varied: Vibrating length L.

Method A — Fixed Frequency

Set the signal generator to a fixed frequency. Then change the vibrating length L until the string resonates. Record the resonant length for several different fixed frequencies.

Set up the apparatus

Attach one end of the string to the vibration generator and pass it over the pulley. Hang the masses on the free end to create tension T. Measure and record T = Mg. This stays constant for every reading.

💡 Add a small pre-tension (at least the hanger alone) before measuring L — this removes any kinks.

Choose and set a fixed frequency

Set the signal generator to a chosen frequency (e.g. 120 Hz). Write it down. Do not change it during this set of readings. Switch the output ON.

💡 Frequencies between 80–400 Hz work well. Start at 120 Hz for string lengths of 0.30–0.80 m.

Adjust L until resonance

Move the bridge or clamp position to change the vibrating length L. Watch the string carefully — at resonance the amplitude suddenly increases and a clear loop pattern appears. The string appears almost stationary.

💡 If you see 2 loops instead of 1, you are at the 2nd harmonic. Increase L (or use half the frequency) to get back to the fundamental.

Fine-tune and measure L

Adjust L slowly around the resonant position to find the length where amplitude is greatest. Measure this length carefully with a metre rule from the vibration generator pin to the pulley. Record L.

💡 Repeat 3 times and take the mean L to reduce random uncertainty. Percentage uncertainty in L is your main source of error.

Change frequency and repeat

Set the signal generator to a new frequency. Find the new resonant length L. Repeat for at least 6 different frequencies across a wide range (e.g. 80, 120, 160, 200, 250, 320 Hz).

💡 A wider range of f (and therefore 1/L) values gives a more reliable gradient on your graph.

Plot f against 1/L

Calculate 1/L for each reading. Plot f (y-axis) against 1/L (x-axis). Draw a best-fit straight line through the origin. The gradient = √(T/μ)/2 = v/2.

💡 The line must pass through the origin — if it doesn't, check for a systematic error in your L measurements (e.g. measuring from the wrong reference point).

Method B — Sweep Frequency

Fix the vibrating length L at a set value. Then sweep the signal generator frequency upward from low until the string resonates. Record the resonant frequency for several different fixed lengths.

Set up the apparatus

Attach the string as before. Apply the hanging masses to set tension T. Clamp the string at a chosen length L (e.g. 0.50 m). Measure L carefully — this is your variable, so accuracy matters.

💡 Mark the clamp position with a pencil mark on the bench so L does not shift during the experiment.

Sweep frequency from low upward

Start the signal generator at a low frequency (e.g. 20 Hz) with the output ON. Slowly increase the frequency. Watch the string — at the resonant frequency the amplitude suddenly becomes much larger and a clear loop pattern appears.

💡 Sweep slowly. If you turn the dial too fast you will pass through resonance without noticing it.

Fine-tune to maximum amplitude

When a resonance appears, adjust the frequency in small steps (±1 Hz) to find the exact frequency where the amplitude is greatest. Read the frequency from the signal generator display. Record f and L.

💡 The resonance indicator in the simulation turns green when you are within ~1% of the true resonant frequency. In a real experiment, watch for the moment the string suddenly vibrates with much greater amplitude.

Note higher harmonics (optional)

Continue sweeping upward — you will find resonance again at f₂ ≈ 2f₁ (2 loops) and f₃ ≈ 3f₁ (3 loops). These all confirm the same f ∝ 1/L relationship and can be added to your graph.

💡 The 2nd harmonic is at exactly twice the fundamental frequency only if the string has uniform μ. Any deviation indicates non-uniformity.

Change L and sweep again

Move the clamp to a new length (e.g. 0.60 m). Repeat the frequency sweep from low to find the new resonant frequency. Repeat for at least 6 different lengths.

💡 Try lengths across a wide range — e.g. 0.20, 0.30, 0.40, 0.50, 0.60, 0.70, 0.80 m. Shorter L gives higher f.

Plot f against 1/L

Calculate 1/L for each reading. Plot f (y-axis) against 1/L (x-axis). Draw a best-fit straight line through the origin. The gradient = √(T/μ)/2 = v/2.

💡 Method B typically gives smaller uncertainty in f than Method A gives in L, making it the preferred method for a precise result.

Fixed Variables

String type Medium nylon

μ (mass/length) 1.20 g/m

Tension T 2.94 N (300 g)

Wave speed v —

Variable — Length L

Vibrating length L 0.50 m

0.10 m1.00 m

Resonance Method

Harmonic

Resonance status

—

Resonant f

—

Length L

—

1/L

—

Harmonic n

—

Readings recorded: 0 / 6 minimum

      Amplitude response curve — ▲ marks current frequency relative to resonance
    

Results Data Table

Fixed: T = 2.94 N, μ = 1.20 g/m. Varied: length L. Derived quantities calculated automatically.

#	n harmonic	L / m	1/L / m⁻¹	f / Hz	f₁ = f/n / Hz	T / N	μ / g m⁻¹
No data yet — go to Simulation tab.

Graph & Analysis — f vs 1/L

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

Uncertainty Estimate

%U in L (ruler ±0.5 mm)—

%U in v = %U in gradient ≈ %U in L—

Dominant uncertainty — L measured with ruler. Half-reading = 0.5 mm.

Discussion Questions

Write your answers and reveal model answers when ready.

Question 1

Explain why the graph of f against 1/L passes through the origin, and what this tells you about the relationship between f and L.

The graph passes through the origin because f = (1/2L)√(T/μ) = k/L where k is a constant. When 1/L = 0 (i.e. L → ∞), f → 0, so the line must pass through the origin. This confirms that f is directly proportional to 1/L (equivalently, f is inversely proportional to L). The relationship f ∝ 1/L means that doubling the vibrating length exactly halves the resonant frequency.

Question 2

The gradient of the f vs 1/L graph equals v/2, where v is the wave speed on the string. Explain why, and show how you can use the gradient to calculate v.

From f = (1/2L)√(T/μ), rearranging: f = (v/2) × (1/L) where v = √(T/μ). This is of the form y = mx with x = 1/L and gradient m = v/2. So the wave speed v = 2 × gradient. Since T and μ are known, the expected wave speed is v = √(T/μ), which can be compared with the measured value to assess experimental accuracy.

Question 3

Explain the difference between the fixed-frequency method and the sweep-frequency method for finding resonance. State one advantage of each.

Fixed-frequency method: the signal generator is set to a constant frequency and the vibrating length L is adjusted until resonance occurs. The measured quantity is L. Advantage: no signal generator display is needed; L can be measured precisely with a ruler. Sweep-frequency method: the length L is fixed and the signal generator frequency is swept upward until resonance occurs. The measured quantity is f. Advantage: f can be read more precisely from a digital display than L from a ruler, reducing percentage uncertainty, especially for short string lengths.

Question 4

A student finds that their f vs 1/L graph does not pass through the origin — the y-intercept is positive. Suggest a reason for this systematic error.

A positive y-intercept means that as 1/L → 0 (L → ∞), f approaches a non-zero value, which is physically impossible. A likely cause is that the student measured L from the wrong reference point — for example, from the vibration generator body rather than the vibrating pin, or from the edge of the pulley rather than its centre. This adds a fixed length δL to every measurement, so the true vibrating length is L + δL rather than L. Using 1/(L + δL) instead of 1/L shifts all x-values to the left and produces a non-zero intercept.

Question 5

State two precautions a student should take to improve the reliability of this experiment.

1. Use the same string throughout — replacing the string changes μ, which changes the resonant frequency for a given L, introducing a systematic error. Measure μ accurately by weighing a 1.00 m length of the string before starting. 2. Measure L consistently from the same reference points — use the tip of the vibration generator pin to the centre of the pulley each time. Any inconsistency in reference points introduces a random or systematic error in L, which directly affects the gradient of the f vs 1/L graph.

CP5b

Vibrating String — Effect of Tension on Resonant Frequency

Edexcel IAL · AS Physics

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.

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.

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).

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.

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.

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.

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.

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.

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.

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.

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.

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.

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).

Uncertainty Estimate

%U in T (from mass ±1g)—

%U in μ from gradient—

T = Mg; %U in T ≈ 2 × %U in M (mass measurement with balance).

Discussion Questions

Write your answers and reveal model answers when ready.

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.

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.

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.

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.

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.

CP5c

Vibrating String — Effect of Mass per Unit Length on Resonant Frequency

Edexcel IAL · AS Physics

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.

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.

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.

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).

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.

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.

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).

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.

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.

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.

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.

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.

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

—

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).

Uncertainty Estimate

%U in μ (mass/length ±0.01 g/m)—

%U in gradient ≈ %U in μ—

Weigh a known length; %U in μ = %U in mass + %U in length.

Discussion Questions

Write your answers and reveal model answers when ready.

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.

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.

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.

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.

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.

Vibrating String

Vibrating String — Effect of Length on Resonant Frequency

Theory — Effect of Length on Frequency

What You Are Investigating

The Key Relationship

Stationary Waves and Harmonics

Fixed Variables in This Experiment

Procedure

Equipment

Results Data Table

Graph & Analysis — f vs 1/L

Uncertainty Estimate

Interpretation

Discussion Questions

Vibrating String — Effect of Tension on Resonant Frequency

Theory — Effect of Tension on Frequency

What You Are Investigating

The Key Relationship

Tension from Hanging Masses

Fixed Variables in This Experiment

Procedure

Equipment

Results Data Table

Graph & Analysis — f vs √T

Uncertainty Estimate

Interpretation

Discussion Questions

Vibrating String — Effect of Mass per Unit Length on Resonant Frequency

Theory — Effect of Mass per Unit Length on Frequency

What You Are Investigating

The Key Relationship

Measuring μ

Fixed Variables in This Experiment

Procedure

Equipment

Results Data Table

Graph & Analysis — f vs 1/√μ

Uncertainty Estimate

Interpretation

Discussion Questions