Lab 5_ Echolation and Acoustic Resonance

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Physics

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Dec 6, 2023

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Part A: Echolocation: 1. Speed of SOUND in air. Assuming T= 22 o C v=332m/s+Tx0.59m/s-°C v=332m/s+(22)(0.59m/s-°C) v =344.93 =345m/s. 2. T= 0.00310s at peak point. 3. V sound = 2de/ Δ T de =VSOUNDX Δ T/2 = (345m/s)x(0.00310s)/2 = 0.535m 4. dt = distance between phone and surface dt= 0.50m 5. Percent difference = |de-dt/dt| x 100% |0.535-0.50/0.50| x 100 = 7%

● Question: Why do you think the best results are obtained when your phone is 0.5 - 1.0 m from the reﬂecting surface? The best results are obtained when the phone is 0.5-1.0m from the reﬂecting surface because the distance is the most important factor in determining the percent difference. If the phone is too close to the surface (less than 0.5m), then even the smallest interference could alter the graph and impact the percent difference by a great amount. If the distance is greater than 1m, there is a greater chance of interference from external factors such as other sounds to disrupt the sound emitted from the phone when doing the experiment. It would be difficult to distinguish between the ambient sounds of the room and the incoming signal the phone is supposed to pick up. ● Question: How precisely could you measure changes in the position of an object using the phyphox echolocation app? Explain how you arrived at your answer. The frequency with which phyphox gathers data points is related to the smallest change in position that may be measured using the phyphox echolocation app. As long as the object's distance stays within the range of 0.5m-1m, it should be possible to measure changes in an object's position as accurately as you can. To ensure the accuracy of this experiment, it is vital to make sure that the only factor affecting the outcome is the item's distance from the phone, the environment around the object and the phone should stay the same. Using the difference between the values of the delay time and the speed of sound in air, the change in distance can be calculated. I arrived at this answer since my percent difference (7%) was low. Previously I tried a length far greater than the distance I used for my final calculation. When I compared these findings, I obtained an accurate measurement that was almost a 60 cm difference between the two distances. This demonstrates that I am capable of measuring the distance between an object and the phone with accuracy.

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