Two identical resonance tubes, each 50 cm long and open at both ends, sound their fundamental notes at 15°C. The speed of sound in air v (m s⁻¹) varies with temperature as v = 331 + 0.6θ, where θ is in °C. If the temperature of one tube is raised to 30°C, how many beats are produced in one second?

2020 A/L Physics Past Paper · Question 21 (English Medium)

A warmed open pipe — beats

Two identical resonance tubes, each $50\ \text{cm}$ long and open at both ends, sound their fundamental notes at $15^\circ\text{C}$ . The speed of sound in air $v\ (\text{m s}^{-1})$ varies with temperature as $v = 331 + 0.6\theta$ , where $\theta$ is in $^\circ\text{C}$ . If the temperature of one tube is raised to $30^\circ\text{C}$ , how many beats are produced in one second?

(1) $4$
(2) $6$
(3) $9$
(4) $12$
(5) $14$

Answer and explanation

Answer: (3) $9$

Given information

We know:

Pipe length $L = 50\ \text{cm} = 0.50$ m, open at both ends.

Starting temperature $15^\circ\text{C}$ ; one pipe is raised to $30^\circ\text{C}$ .

$v = 331 + 0.6\theta$ ; both pipes sound their fundamental notes.

What the question asks

We want: the beats per second $f_b$ when the two pipes sound together.

Relevant theory

The relationship among the wave velocity v, frequency f and the wavelength λ, can be written in the form of v = f λ.

The number of times the sound reaches maximum in one second is called beat frequency,

Required equations

$\lambda_1 = 2L$

$f = \frac{v}{\lambda_1} = \frac{v}{2L}$

$f_b = |f_1 - f_2|$

Applying it to this question

Plan:

Find the speed of sound at both temperatures.

Find each pipe’s fundamental note $v/2L$ .

Their difference is the beats.

The end correction $e$ is not given, so it is neglected.

Reasoning and calculation

$340$

$v_1 = 331 + 0.6 \times 15$

$349$

$v_2 = 331 + 0.6 \times 30$

$f_1 = \frac{340}{2 \times 0.50}$

$f_2 = \frac{349}{2 \times 0.50}$

$9$

$f_b = 349 - 340$

Checks

Units: $\dfrac{\text{m s}^{-1}}{\text{m}} = \text{s}^{-1} = \text{Hz}$ ; beats are also per second ✓.

Does it make sense? Since $2L = 1$ m, every $1\ \text{m s}^{-1}$ change in speed is $1$ Hz; $0.6 \times 15 = 9\ \text{m s}^{-1}$ ⇒ $9$ beats. Nine beats per second is something an ear could count ✓.

$0.6 * (t2 - t1) / (2 * L)$

Conclusion

$f_b = 349 - 340 = 9\ \text{s}^{-1}$

$f_1 = 340$ Hz, $f_2 = 349$ Hz; $f_b = 9$ — the answer is (3).

Using the closed-pipe formula $f = \dfrac{v}{4L}$ for an open pipe.

Both ends are open; the fundamental has an antinode at each end: $\lambda = 2L$ .

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