Wave Speed on a Stretched Spring (Edexcel International A Level Physics)

Revision Note

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Lindsay Gilmour

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Physics

Wave Speed on a Stretched String

  • The speed of a wave travelling along a string with two fixed ends is given by:

Velocity Equation

  • Where:
    • T = tension in the string (N)
    • μ = mass per unit length of the string (kg m–1)

 

  • At the fundamental frequency, f0 of a stationary wave of length L, the wavelength, λ = 2L
  • Therefore, according to the wave equation, the speed of the stationary wave is:

v = = f × 2L

  • Combining these two equations leads to the equation for the fundamental frequency (sometimes referred to as the first harmonic):

Frequency of First Harmonic

  • Where:
    • f = frequency (Hz)
    • L = the length of the string (m)
    • = the tension in the string (N)
    • µ = mass per unit length (kg m-1)

  • Mass per unit length, µ can be calculated by dividing the mass of the string by the length of the string

Fixed end wavelengths and harmonics (1), downloadable AS & A Level Physics revision notesFixed end wavelengths and harmonics (2), downloadable AS & A Level Physics revision notes

Diagram showing the first three modes of vibration of a stretched string with corresponding frequencies

Worked example

A guitar string of mass 3.2 g and length 90 cm is fixed onto a guitar.

The string is tightened to a tension of 65 N between two bridges at a distance of 75 cm.

Guitar String First Harmonic Question, downloadable AS & A Level Physics revision notesCalculate the

a) speed of the waves on the string

b) fundamental frequency of the string

Part (a)

Step 1: Write the known quantities in S.I. units

    • Tension, T = 65 N
    • Mass, m = 3.2 g = 3.2 × 10−3 kg
    • Length of string, L = 90 cm = 0.90 m
      • Mass per unit length, μ m over L equals fraction numerator 3.2 cross times 10 to the power of negative 3 end exponent over denominator 0.9 end fraction = 3.56 × 10−3 kg m−1

Step 2: Write the equation for speed on a string and calculate

    • v = = f × 2L AND fraction numerator 1 over denominator 2 L end fraction square root of T over mu end root
    • So, v = fraction numerator 1 over denominator 2 L end fraction square root of T over mu end root cross times 2 L space equals space square root of T over mu end root

v equals square root of T over mu end root space equals square root of fraction numerator 65 over denominator 3.56 space cross times space 10 to the power of negative 3 end exponent end fraction end root space = 135

Step 3: Write the answer to the correct significant figures and include units

    • The speed of the wave on the string, v = 140 m s−1

Part (b)

Step 1: Write the known quantities in S.I. units

    • Tension, T = 65 N
    • Length of string under tension, L = 75 cm = 0.75 m
    • Mass per unit length, μ = 3.56 × 10−3 kg m−1 (from part (a))

Step 2: Identify the length of one wavelength at the fundamental frequency, f0

Guitar String First Harmonic Answer

Step 3: Write the equation for fundamental frequency and calculate

f subscript 0 space equals space fraction numerator 1 over denominator 2 L end fraction square root of T over mu end root space equals space fraction numerator v over denominator 2 L end fraction space equals space fraction numerator 135 over denominator left parenthesis 2 space cross times space 0.75 right parenthesis end fraction = 90.1

Step 3: Write the answer to the correct significant figures and include units

    • The fundamental frequency, f0 = 90 Hz

Exam Tip

Go through solutions step by step, showing all your working. Questions like this one will be very similar so you can rely on using your tried and tested method to get the answer.

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Lindsay Gilmour

Author: Lindsay Gilmour

Lindsay graduated with First Class Honours from the University of Greenwich and earned her Science Communication MSc at Imperial College London. Now with many years’ experience as a Head of Physics and Examiner for A Level and IGCSE Physics (and Biology!), her love of communicating, educating and Physics has brought her to Save My Exams where she hopes to help as many students as possible on their next steps.

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