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Edexcel A Level (IAL) Physics-2.11 Core Practical 5: Investigating Stationary Waves- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -2.11 Core Practical 5: Investigating Stationary Waves- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -2.11 Core Practical 5: Investigating Stationary Waves- Study Notes -Edexcel A level Physics – per latest Syllabus.

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Edexcel A level Physics-Study Notes- All Topics

CORE PRACTICAL 5: Investigate How Length, Tension and Mass per Unit Length Affect the Frequency of a Vibrating String

This practical investigates how the frequency of a vibrating string depends on: (1) its length, (2) its tension, and (3) its mass per unit length. A signal generator and vibration transducer are used to excite the string.

Aim

To measure how the frequency of the fundamental mode (first harmonic) of a vibrating string changes with:

  • string length \( L \)
  • tension \( T \)
  • mass per unit length \( \mu \)

The theoretical relationship is:

$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$

Apparatus

  • String or wire
  • Signal generator
  • Vibration transducer / mechanical oscillator
  • Pulley, masses and hanger (to apply tension)
  • Metre ruler
  • Micrometer (to find mass per unit length if required)
  • Digital balance
  • Fixed bridge/clamps

Theory

The fundamental frequency of a stretched string is:

$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $

  • Increasing tension → higher frequency.
  • Increasing length → lower frequency.
  • Increasing mass per unit length → lower frequency.

This relationship is derived from wave speed:

$ v = \sqrt{\frac{T}{\mu}}, \qquad f = \frac{v}{2L} $

Procedure (General)

  1. Fix one end of the string and pass the other end over a pulley with masses attached (to control tension).
  2. Attach a vibration transducer near the fixed end and connect it to the signal generator.
  3. Set the string into vibration using the signal generator.
  4. Adjust the frequency until the first harmonic is observed (one antinode in the middle).
  5. Record the frequency.
  6. Repeat the experiment while varying:
    • Length \( L \)
    • Tension \( T \)
    • Mass per unit length \( \mu \)
  7. Take several readings and calculate averages.

Part A: Effect of Length

  1. Keep tension \( T \) and mass per unit length \( \mu \) constant.
  2. Change the vibrating length by moving bridges along the string.
  3. Find the fundamental frequency for each length.
  4. Plot:

    \( f \) on y-axis vs \( \frac{1}{L} \) on x-axis.

  5. A straight line should be obtained.

Part B: Effect of Tension

  1. Keep length \( L \) and mass per unit length \( \mu \) constant.
  2. Change tension by adding masses to hanger.
  3. Record frequency for each tension.
  4. Plot:

    \( f \) vs \( \sqrt{T} \)

  5. A straight line should be obtained.

Part C: Effect of Mass per Unit Length

  1. Use strings or wires of different mass per unit length \( \mu \).
  2. Keep tension and length constant.
  3. Find frequency for each string.
  4. Plot:

    \( f \) vs \( \frac{1}{\sqrt{\mu}} \)

  5. A straight line should be obtained.

Sources of Error

  • Difficulty identifying exact resonance frequency.
  • Vibrations not in the fundamental mode.
  • Air resistance or damping.
  • String not perfectly horizontal.
  • Mass hanger friction in pulley.
  • String stiffness (real strings aren’t perfectly flexible).

Improvements

  • Use a well-calibrated digital signal generator.
  • Take readings several times and average.
  • Ensure string is straight and parallel to bench.
  • Use heavier masses to reduce percentage uncertainty in tension.
  • Use a low-mass, low-friction pulley.

Example Calculation

A string of length \( L = 0.80\,\mathrm{m} \), tension \( T = 16\,\mathrm{N} \), and mass per unit length \( \mu = 0.004\,\mathrm{kg\,m^{-1}} \) vibrates in its fundamental mode. Calculate the frequency.

▶️ Answer / Explanation

Step 1: Use the formula

$ f = \frac{1}{2L}\sqrt{\frac{T}{\mu}} $

Step 2: Substitute values

$ f = \frac{1}{2(0.80)}\sqrt{\frac{16}{0.004}} $

$ = \frac{1}{1.6}\sqrt{4000} = 0.625 \times 63.25 \approx 39.5\,\mathrm{Hz} $

Frequency ≈ \( 40\,\mathrm{Hz} \)

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