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Edexcel A Level (IAL) Physics-2.10 Wave Speed on a Stretched Spring- Study Notes- New Syllabus

Edexcel A Level (IAL) Physics -2.10 Wave Speed on a Stretched Spring- Study Notes- New syllabus

Edexcel A Level (IAL) Physics -2.10 Wave Speed on a Stretched Spring- Study Notes -Edexcel A level Physics – per latest Syllabus.

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Edexcel A level Physics-Study Notes- All Topics

Speed of a Transverse Wave on a String \( v = \sqrt{\dfrac{T}{\mu}} \)

This formula gives the speed of a transverse wave travelling along a stretched string or wire. It depends on how tight the string is and how much mass it has per unit length.

The Wave Speed Formula

The speed of a transverse wave on a string is:

$ v = \sqrt{\frac{T}{\mu}} $

  • \( v \) = wave speed (m s⁻¹)
  • \( T \) = tension in the string (N)
  • \( \mu \) = mass per unit length of the string (kg m⁻¹)

Meaning:

  • Increasing tension → increases wave speed.
  • Increasing mass per unit length → decreases wave speed.
  • Speed does NOT depend on the amplitude or frequency of the oscillations.

Understanding Mass per Unit Length \( \mu \)

\( \mu \) tells us how “heavy” a string is for each metre of length. It is calculated by:

$ \mu = \frac{m}{L} $

  • \( m \) = mass of the string (kg)
  • \( L \) = length of the string (m)

Thicker or denser strings have larger \( \mu \) → waves travel more slowly.

Physical Interpretation

  • A tighter string pulls the wave forward faster → larger \( T \).
  • A heavier string resists acceleration → larger \( \mu \).
  • The formula resembles Newton’s second law: higher tension provides more force; higher mass per metre resists motion.

 Applications

  • Vibrating strings on musical instruments.
  • Laboratory experiments for measuring tension & string density.
  • Understanding harmonics and standing waves on strings.
  • Stringed resonance tubes in physics experiments.

Using the Formula Correctly

  • Always convert mass of string to kilograms.
  • Ensure \( \mu \) is in kg/m.
  • Tension must be measured in newtons.
  • Wave speed depends only on string properties — not on the driving frequency.

Example (Easy)

A string has tension \( 20\,\mathrm{N} \) and mass per unit length \( 0.010\,\mathrm{kg\,m^{-1}} \). Find the wave speed.

▶️ Answer / Explanation

$ v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{20}{0.010}} = \sqrt{2000} \approx 44.7\,\mathrm{m\,s^{-1}} $

Example (Medium)

A 3.0 m long string has a mass of 0.12 kg and is under 30 N tension. Find the speed of a wave on the string.

▶️ Answer / Explanation

Step 1: Find mass per unit length

$ \mu = \frac{m}{L} = \frac{0.12}{3.0} = 0.040\,\mathrm{kg\,m^{-1}} $

Step 2: Use wave speed formula

$ v = \sqrt{\frac{30}{0.040}} = \sqrt{750} \approx 27.4\,\mathrm{m\,s^{-1}} $

Example (Hard)

A string vibrates with a wave speed of \( 65\,\mathrm{m\,s^{-1}} \) when the tension is \( 50\,\mathrm{N} \). Calculate the mass per unit length of the string.

▶️ Answer / Explanation

Use the formula rearranged:

$ v = \sqrt{\frac{T}{\mu}} \quad\Rightarrow\quad \mu = \frac{T}{v^2} $

$ \mu = \frac{50}{65^2} = \frac{50}{4225} \approx 0.0118\,\mathrm{kg\,m^{-1}} $

Mass per unit length = \( 1.18\times10^{-2}\,\mathrm{kg\,m^{-1}} \)

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