9702 · 8.1

Stationary waves — practice questions

Practice and worked examples for 9702 Stationary waves. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

A string of length 0.80 m has a mass of 4.0 g. When stretched with a tension of 20 N, calculate the fundamental frequency of vibration.

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First, calculate the mass per unit length, $\mu$ . Remember to convert mass to kilograms. $\mu = \text{mass} / \text{length} = 0.004 \text{ kg} / 0.80 \text{ m} = 0.005 \text{ kg m}^{-1}$ .
Now, apply the formula for the fundamental frequency ( $f$ ). $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$
Substitute the given values into the formula: $f = \frac{1}{2 \times 0.80} \sqrt{\frac{20}{0.005}}$
Calculate the result: $f = \frac{1}{1.6} \sqrt{4000}$ $f \approx 0.625 \times 63.2455...$ $f \approx 39.5 \text{ Hz}$

Worked example 2

A guitar string of length 75 cm is fixed at both ends. It is plucked and vibrates in its third harmonic mode. The frequency of this sound is found to be 660 Hz. (a) Determine the wavelength of the waves on the string. (b) Calculate the speed of the progressive waves on the string.

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First, relate the string length to the wavelength for the third harmonic. The third harmonic (n=3) has three antinodes ('loops') on the string. The total length L contains three half-wavelengths. $L = 3 \times (\frac{\lambda}{2})$
Rearrange the formula and substitute the values to find the wavelength (λ). Remember to convert length to metres. $L = 0.75 \text{ m}$ $\lambda = \frac{2L}{3} = \frac{2 \times 0.75}{3}$ $\lambda = \frac{1.50}{3} = 0.50 \text{ m}$ The wavelength is 0.50 m.
Now, use the wave speed equation, $v = f\lambda$ , to find the speed of the waves. $v = 660 \text{ Hz} \times 0.50 \text{ m}$ $v = 330 \text{ m s}^{-1}$ The speed of the waves on the string is 330 m s⁻¹.

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