3.1.7

# Stationary Waves 2

Test yourself

## Stationary Wave Practical

This practical investigates how the frequency of stationary waves on a string changes when length, tension and mass per unit length are changed.

### Initial measurements

• Measure the length of the string, the tension in the string and the mass per unit length.
• The tension can be calculated by multiplying the total mass of the masses and the acceleration due to gravity.
• $T = mg$
• The mass per unit length can be calculated by dividing the total mass of the string by its length.
• $\mu = \frac{M}{l}$

### Determine frequency of first harmonic

• This can be found by varying the frequency until you see the pattern needed for the first harmonic.
• Two nodes at each end of the string with one antinode in the centre.

### Change variable - length

• To investigate how changing the length affects the resonant frequency, keep the tension and mass per unit length the same.
• Vary the length of the string by moving the oscillator away or toward the pulley.
• At each string length, find the new first harmonic.
• Plot a graph of f against l.

### Change variable - mass per unit length

• To investigate how changing μ affects the resonant frequency, keep the tension and length the same.
• Vary μ by using different types of material for the string.
• For each material (with a different value of μ), find the new first harmonic.
• Plot a graph of μ against f.

### Change variable - tension

• To investigate how changing tension affects the resonant frequency, keep the length and mass per unit length the same.
• Vary the tension in the string by varying the mass attached to the end.
• At each value of tension, find the new first harmonic.
• Plot a graph of f against T.

## Stationary Wave Practical - Results

The frequency of the first harmonic of a string can be calculated using an equation. The practical shown previously should show certain relationships.

### First harmonic equation

• We expect that the frequency of the first harmonic should be given by:
• $f = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$

### Relationship with length

• The longer the length of the string, the lower the frequency.
• This is because the longer the length is, the longer the half wavelength must be, and so the lower the frequency.

### Relationship with μ

• The larger the mass per unit length, μ, the lower the resonant frequency.
• For the same length string, waves travel more slowly through a heavier string, so the frequency must be lower.
• Remember $f = \frac{v}{\lambda}$

### Relationship with tension

• The higher the tension in the string, the higher the frequency.
• This is because waves travel more quickly down a string with higher tension.