KEY TAKEAWAY: Wave superposition is the fundamental principle behind the formation of standing waves.
STUDY HINT: Visualize the waves traveling back and forth and interfering to create the standing wave pattern.
APPLICATION: Musical instruments rely on standing waves to produce sound. Different harmonics create different tones or timbres.
For a string of length $L$ fixed at both ends, the possible wavelengths ($\lambda_n$) for standing waves are given by:
$$L = n \frac{\lambda_n}{2}$$
where:
Rearranging for $\lambda_n$:
$$\lambda_n = \frac{2L}{n}$$
The frequency ($f_n$) of the $n$th harmonic is related to the wave velocity ($v$) and wavelength ($\lambda_n$) by:
$$v = f_n \lambda_n$$
Substituting $\lambda_n = \frac{2L}{n}$:
$$f_n = \frac{nv}{2L}$$
where:
The fundamental frequency ($f_1$) is the lowest resonant frequency and corresponds to $n = 1$:
$$f_1 = \frac{v}{2L}$$
The frequencies of higher harmonics are integer multiples of the fundamental frequency:
$$f_n = nf_1$$
| Harmonic (n) | Frequency ($f_n$) | Wavelength ($\lambda_n$) | Number of Antinodes | Number of Nodes |
|---|---|---|---|---|
| 1 | $f_1$ | $2L$ | 1 | 2 |
| 2 | $2f_1$ | $L$ | 2 | 3 |
| 3 | $3f_1$ | $2L/3$ | 3 | 4 |
| 4 | $4f_1$ | $L/2$ | 4 | 5 |
REMEMBER:
L = n(lambda/2)- helps recall the length and wavelength relationship.
The speed of a transverse wave on a string is determined by the tension ($T$) in the string and the linear mass density ($\mu$) of the string:
$$v = \sqrt{\frac{T}{\mu}}$$
where:
COMMON MISTAKE: Forgetting to convert units (e.g., cm to m, grams to kg) when calculating wave speed or frequency.
A guitar string of length 0.65 m is fixed at both ends. The speed of the wave on the string is 195 m/s. Calculate the fundamental frequency and the frequency of the second harmonic.
$$f_1 = \frac{v}{2L} = \frac{195}{2 \times 0.65} = 150 \text{ Hz}$$
$$f_2 = 2f_1 = 2 \times 150 = 300 \text{ Hz}$$
A string of length 1.5 m and mass 0.015 kg is fixed at both ends. The string is under a tension of 45 N. Calculate the speed of the wave on the string and the fundamental frequency.
$$\mu = \frac{m}{L} = \frac{0.015}{1.5} = 0.01 \text{ kg/m}$$
$$v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{45}{0.01}} = \sqrt{4500} = 67.08 \text{ m/s}$$
$$f_1 = \frac{v}{2L} = \frac{67.08}{2 \times 1.5} = 22.36 \text{ Hz}$$
EXAM TIP: Practice applying the formulas for wavelength, frequency, and wave speed in different scenarios. Pay attention to units and significant figures.
VCAA FOCUS: VCAA often tests the relationships between length, tension, linear mass density, and the resulting frequencies of standing waves. They might provide scenarios where one of these parameters is changed and ask how the frequency is affected.
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