Part 3: Measure speed of sound In this part of the lab yau tracked a single peak

Question

Part 3: Measure speed of sound In this part of the lab yau tracked a single peak as you moved a microphone in order to get a good value of the speed of sound. This question will lead you through a similar process with just two measurements, The graphs below show two sine-wave signals like you saw from a microphone positioned in front of a speaker in Part 3 of the experiment. The two dashed vertical lines represent the cursors of the oscilloscope, with the one on the right side (colored orange) tracking peak of the sound wave as the microphone is moved. The position x of the microphone in front of the speaker and the time between the cursors is shown below each graph. 0.4 0.2 0.0 0.2 0.4 0.0 0.2 0.4 0.6 0.8 1.0 Time (ms Mic position: x-5.00 cm. Cursor difference: ar-0.123 ms 0.4 0.2 0.0 -0.2 -0.4 0.0 0.2 0.4 0.6 0.8 1.0 Time (ms) Mic position: x = 20.0 cm. Cursor difference: = 0.552 ms For all entries, your answers must be correct to within 2%. 1. What is the speed of sound indicated by the above graphs? v= 0.3375 micr X 2. In the top graph the phase difference between the two cursors is /2 radians. Based on this and the information given what is The frequency f = The wavelength = 2 Hz 2 m 3. In Steps 3.6-3.8 the claim is made that the speed of sound in air depends on temperature. The reason for this is that temperature affects the density of the gas (higher temperature means lower density) and this lower density causes the speed to increase. This is similar to the phenomenon that lighter guitar ar violin strings have a higher pitth than the heavier ones. The specific theory of sound propagation in gases goes beyond the topics of this course, but it is interesting to nate that there is a relationship between the sound speed and the temperature, and one could use a sound measurement, in principle, to obtain the temperature.

Explanation / Answer

a. from the given graphs

speed of sound = (difference in position of microphones)/( differnce in position of cursor)

now given x1 = 5 cm, t1 = 0.123 ms

x2 = 20 cm, t2 = 0.552 ms

hence

speed of sound c = (x2 - x1)/(t2 - t1) = (0.2 - 0.05)/(0.552 - 0.123)*10^-3 = 349.65 m/s

b. given that the phase difference between two cursors is pi/2

so, wavelength = 2(x2 - x1) = 2(0.2 - 0.05) = 0.4 - 0.1 = 0.3 m

frequency = f

then

f*lambda = c

f = 349.65/0.3 = 1165.5011 Hz

c. we know that relationship between the temperature and speed of sound in medium it

c = sqroot(gamma*RT/M)

gamma for air = 1.4

R = 8.31

hence

Molar mass of air M = 0.029 kg/mol

349.65^2 = 1.4*8.31*T/0.029

T = 304.744 K = 31.58 C

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