A student sets up an experiment to measure the speed of sound in air accurately. A loudspeaker is connected to a signal generator that produces a sound of f...
Question 1 Report
A student sets up an experiment to measure the speed of sound in air accurately. A loudspeaker is connected to a signal generator that produces a sound of frequency 1700 Hz. Two microphones M1 and M2 are placed along a straight line from the loudspeaker. M1 is 0.50 m from the loudspeaker and M2 is 1.50 m from the loudspeaker. Both microphones are connected to an electronic timer that measures the time interval between the sound arriving at M1 and at M2. Fig. 1.1 shows the arrangement. The electronic timer records a time of 2.94 × 10−3 s. The student then connects M1 to a cathode-ray oscilloscope with the time base set to 0.10 ms per division. The experiment is performed in a large hall at room temperature to minimise reflections from nearby walls.
Fig. 1.1
(a) Calculate the distance between the two microphones. [1]
(b) Calculate the speed of sound in air. [2]
(c) State one advantage of using electronic timing rather than a stopwatch for this experiment. [1]
(d) Determine the period of the 1700 Hz sound wave. [1]
(e) Calculate the number of divisions occupied by one complete cycle on the CRO screen. [2]
(f) Calculate the wavelength of the sound in air. [2]
(g) State the type of wave produced by the loudspeaker in air. [1]
Answer Details
(a) Calculate the distance between the two microphones. [1]
M1 is 0.50 m from the loudspeaker and M2 is 1.50 m from the loudspeaker. Since both are on the same straight line:
\(d = 1.50 - 0.50 = 1.00\text{ m}\) [1]
The timer measures the time for sound to travel this 1.00 m gap, not the full distance from the speaker. Using two microphones eliminates uncertainty about exactly when the speaker emits the pulse.
(b) Calculate the speed of sound in air. [2]
\(v = \frac{d}{t} = \frac{1.00}{2.94 \times 10^{-3}}\) [1]
\(v = 340\text{ m/s}\) [1]
This is consistent with the accepted value for the speed of sound in air at room temperature (~340 m/s).
(c) State one advantage of using electronic timing rather than a stopwatch for this experiment. [1]
Electronic timing eliminates human reaction time error. The time interval measured (about 3 ms) is far too short for a human to start and stop a stopwatch accurately. Electronic sensors respond to the sound arrival with much greater precision and consistency. [1]
(d) Determine the period of the 1700 Hz sound wave. [1]
\(T = \frac{1}{f} = \frac{1}{1700} = 5.88 \times 10^{-4}\text{ s}\) (0.588 ms) [1]
The period is the time for one complete oscillation. Since frequency and period are reciprocals, a higher frequency means a shorter period.
(e) Calculate the number of divisions occupied by one complete cycle on the CRO screen. [2]
The time base is set to 0.10 ms per division. One cycle takes 0.588 ms:
\(\text{divisions} = \frac{T}{\text{time base}} = \frac{0.588}{0.10}\) [1]
\(= 5.9\text{ divisions}\) (approximately 6 divisions) [1]
(f) Calculate the wavelength of the sound in air. [2]
Using the wave equation \(v = f\lambda\):
\(\lambda = \frac{v}{f} = \frac{340}{1700}\) [1]
\(\lambda = 0.20\text{ m}\) [1]
(g) State the type of wave produced by the loudspeaker in air. [1]
Longitudinal. [1] Sound in air is always a longitudinal wave because the air particles vibrate parallel to the direction of energy transfer, creating compressions and rarefactions along the wave's path.
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