Not sure how to start this problem, any help is appriciated. The graphs below sh

Question

Not sure how to start this problem, any help is appriciated.

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 a 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 Delta t is shown below each graph. For all entries, your answers must be correct to within 2%. What is the speed of sound indicated by the above graphs? In the top graph the phase difference between the two cursors is nil radians. Based on this and the information given what is 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 or violin strings have a higher pitch than the heavier ones. The specific theory of sound propagation in gases goes beyond the topics of this course, but it is interesting to note 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. With this in mind, use the equation above Step 3.6 to find the estimated temperature of the gas in this simulated data. You will need to do a little algebra to solve that equation for the temperature given the speed v.

Explanation / Answer

(1) speed of sound v = (change in distance) / (change in time)

v = (20.0 cm - 5 cm) / (0.531 ms - 0.0828 ms)

v = (15 cm) / (0.4482 ms)

v = (0.15 m) / (0.0004482 s)

v = 334.67 m/s

(2) From the top graph, since phase difference between two cursors is /2, this means the time corresponding to phase of /2 multiplied by 4 will give us the time period
T = 4 * t = 4 * 0.0828 ms = 0.3312 ms = 0.0003312 s

Frequency f = (1/T) = (1 / 0.0003312) = 3019.32 Hz

Wavelength = v / f = (334.67 m/s / 3019.32 Hz) = 0.11 m

(3) For this part, I need the equation in step 3.6 which is mentioned in the question but not shown

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