Question

In an experiment to measure the speed of sound in air, standing waves are set up in a narrow pipe open at both ends using a speaker driven at \(702 Hz\). The length of the pipe is \(2.0 \mathrm{m} .\) What is the air temperature inside the pipe (assumed reasonably near room temperature, \(20^{\circ} \mathrm{C}\) to \(\left.35^{\circ} \mathrm{C}\right) ?\) [Hint: The standing wave is not necessarily the fundamental.]

Short answer

Answer: The approximate air temperature inside the pipe is 25.33°C.

Step-by-step solution

Set up the equation for the speed of sound in air.

In order to find the air temperature, we need to start with the formula for the speed of sound in air in terms of the temperature (in Celsius): \(v = 331.4 + 0.6 T\) where \(v\) is the speed of sound in m/s, and \(T\) is the temperature in Celsius.

Calculate the wavelength of the standing wave.

Given the frequency of the speaker (f = 702 Hz), it's possible to calculate the wavelength (λ) by using the relationship between speed, frequency, and wavelength: \(v = f \cdot \lambda\) Knowing the length of the pipe (L = 2.0 m) and considering that it is open at both ends, the wavelength of standing waves can be expressed as: \(\lambda_n = \frac{2L}{n}\) Where \(n\) is the harmonic number (1, 2, 3, ...).

Express the speed of sound in terms of harmonic number.

Now, we will express the speed of sound and temperature using the harmonic number: \(v_n = f \cdot \lambda_n = f \cdot \frac{2L}{n}\) Since we don't know the exact harmonic number, we'll first need to calculate the possible values of n.

Calculate candidate values for the harmonic number.

First, we need to calculate the v for the given range of temperatures (assuming room temperature of \(20^\circ C\), to \(35^\circ C\)). At \(20^\circ C\), \(v_{20} = 331.4 + 0.6(20) = 343.4 \mathrm{m/s}\) At \(35^\circ C\), \(v_{35} = 331.4 + 0.6(35) = 353.4 \mathrm{m/s}\) Next, we will calculate the possible values of the harmonic number for this range of speed. \(n_{lower} = \frac{2L}{v_{20}} \cdot f = \frac{2(2)}{343.4} \cdot 702 \approx 10.27\) \(n_{higher} = \frac{2L}{v_{35}} \cdot f = \frac{2(2)}{353.4} \cdot 702 \approx 9.96\) Since the harmonic number is an integer, we can conclude that the closest integer value is \(n = 10\).

Calculate the speed of sound and corresponding air temperature.

Now that we have found the possible harmonic number (n = 10), we can use it to find the speed of sound: \(v_{10} = f \cdot \frac{2L}{10} = 702 \cdot \frac{2(2)}{10} = 280.8\ \mathrm{m/s}\) Now, we can use the speed of sound (v = 280.8 m/s) to find the air temperature inside the pipe: \(T = \frac{v - 331.4}{0.6} = \frac{280.8 - 331.4}{0.6} \approx 25.33^\circ C\) Thus, the air temperature inside the pipe is approximately \(25.33^\circ C\).