Fig. 1 shows two divers communicating underwater using a signal device. Diver A strikes a metal plate to make a sound pulse. Diver B receives the pulse 0.20...
Question 1 Report
Fig. 1 shows two divers communicating underwater using a signal device. Diver A strikes a metal plate to make a sound pulse. Diver B receives the pulse 0.20 s later at a distance of 300 m. The divers cannot rely on ordinary speech over this distance because the equipment masks their voices and the water is noisy. The diagram shows the pulse travelling directly through seawater. The divers compare this result with sound travelling through air. Their instructor reminds them that wave velocity depends on the medium, while frequency is set by the source.
(a) Calculate the velocity of sound in the seawater. [2]
(b) Explain why sound can travel through seawater. [2]
(c) What happens to the frequency of the pulse when it enters the water from the metal plate? [1]
(d) Give one reason why a short pulse is useful for communication. [1]
Answer Details
(a) Use \(v=\frac{d}{t}\):
\[v=\frac{300\ \text{m}}{0.20\ \text{s}}=1500\ \text{m s}^{-1}\]Velocity of sound in seawater = \(1500\ \text{m s}^{-1}\). [2]
(b) Seawater contains particles. [1] The particles vibrate and pass the disturbance, or energy, to neighbouring particles. [1] Sound is therefore able to travel through it.
(c) The frequency stays the same. [1] Frequency is set by the source; changing medium changes wave speed and wavelength instead.
(d) A short pulse has a clear start, separate pulses are less likely to overlap, or its arrival time can be measured clearly. [1]
Download The App On Google Playstore
Everything you need to excel in your exams
