Question 1 Report
The diagram in Fig. 1 is an oscilloscope trace made by a student recording a note from a violin. The horizontal scale represents time. The student keeps the microphone at the same distance from the violin for each recording. She compares this note with a second violin note whose trace has more waves in the same time interval. Both notes travel through air to the microphone. She is investigating why one musician can make a note sound higher without making it noticeably louder. The oscilloscope is connected by a cable to the microphone in a simple circuit.
(a) What is the name of the instrument used to display this trace? [1]
(b) Which trace has the greater frequency: Fig. 1 or the second trace? [1]
(c) Draw, on the diagram, one more wave that has the same amplitude and wavelength as those shown. [2]
Fig. 1 shows an underground survey team using an echo-sounding device above a flooded tunnel. The device sends a short pulse of sound through water towards the tunnel floor. The pulse returns 0.040 s after it is sent. The speed of sound in water is 1500 m/s. The measured time includes the journey down to the floor and back to the device. A worker keeps the equipment inside a waterproof box and records the depth before moving the pump. The team uses the result to decide whether a person can enter safely. Ignore any sound reflected from the tunnel walls.
(a) Calculate the total distance travelled by the sound pulse. [2]
(b) Calculate the depth of the water above the tunnel floor. [1]
(c) What name is given to the returning sound pulse? [1]
(d) Describe why the calculated depth is half the total distance travelled. [1]
Oscilloscope trace
(a) The instrument used to display the trace is an oscilloscope (or cathode-ray oscilloscope). [1]
(b) The second trace has the greater frequency because it contains more waves in the same time interval. [1]
(c) The continuation must contain one complete wave [1] with the same amplitude and wavelength as the existing trace. [1]
Echo sounding
(a) \[\text{distance}=\text{speed}\times\text{time}=1500\text{ m/s}\times0.040\text{ s}=60\text{ m}\]
The total distance travelled is 60 m. [2]
(b) The pulse travels down and back, so the water depth is \(\frac{60}{2}=\) 30 m. [1]
(c) The returning pulse is an echo, or reflected pulse. [1]
(d) The depth is half the total distance because the pulse travels to the floor and then back to the device. [1]
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