Did You Calculate the Speed of Sound Correctly?

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Homework Help Overview

The discussion revolves around calculating the speed of sound in a scenario involving sound waves traveling between two points, with reflections off a wall. Participants are analyzing equations and attempting to clarify the relationships between distance, time, and speed.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss various equations related to the problem, questioning the significance of certain variables and the setup of the scenario. There are attempts to clarify the meaning of the distance 'd' and the paths of sound signals. Some participants express confusion over the calculations leading to different speed values.

Discussion Status

The conversation includes multiple interpretations of the problem, with participants providing observations and suggestions for clarity. Some guidance has been offered regarding the setup of the problem and the equations used, but there is no explicit consensus on the correct approach or solution.

Contextual Notes

Participants note discrepancies in the calculated speed of sound, with references to specific values and the implications of time differences in the context of the problem. There is mention of a potential mistake in the original calculations and the need for clearer diagrams to represent the scenario accurately.

Creed
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Homework Statement
Two persons, A and B were interest to find out the speed of the sound in the air. They stand 40 m of a wall and 60 m from each other. The person B hears the sound that A said. And after 1/8 seconds he hears the echo created by the wall. Based on this measures, what is the value these two people obtained for the speed of sound?

Solution: 340 m/s
Relevant Equations
v = d/t
iiii10.png

d² = 40² + 60² ---> I

60 = V.t ---> II

40 + d = V.(t + 1/8) ---> III

Thanks in the advance. :)
 
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If A wanted to throw a ball such that it went to B by first bouncing off the wall, at what point would the ball need to hit the wall?
 
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In the middle of it.
 
I can see a reason for I and II, despite your lack of explanation, but not for III.
Maybe you can mark d on your diagram. I only know where d is from eqn I. I do not know the significance of d.

I wonder what the significance of the grey arrows is? B is looking at A and A is looking at the wall??
I think they are not the paths of the two sound signals.
Perhaps you should show the paths of the two sound signals.

But your approach is right. I think you need to be clearer about what's happening.
Also I can't get 340 m/s by your idea nor mine, so I don't know what they mean by saying it is the solution.

Edit: To get a speed of 340 m/sec when the difference in time is 1/8 sec, they would need a path difference of (1/8) sec x 340 m / sec = 42.5 m
 
Last edited:
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Thanks for helping out, Merlin 3189 and Tsny. I really appreciate it.

Well observed.

There's a mistake, the solution is 320 m/s. The grey arrows should have the end curved in both sides, so A is looking at B and B looking at A and A is looking at wall and wall is facing A. I'm sorry for the draw I had to made at the computer with limited resources.

I just solved the exercise.

With different approach using what Tsny asked me. We have to considered the sound wave bouncing back in the middle of the wall. And using the Pythagorean theorem to find out the distance of wave goes.
That is 50 m from A to middle point of the wall plus 50m from middle point of the wall to B, 100m in Total.

So if, first sound wave travels 60 m.
The second sound wave that goes to A to the middle point of the wall also travels 60 m. Therefore, there is only 40 m left to travel.

Therefore:

v = d/t <=> v = 40 / (1/8)<=> v=320 m/s.

Thank you again to both you.
 
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I did also wonder why you didn't get an answer from your 3 equations:
I ⇒ d =√5200 = 72.11
III ⇒ 40+72.11 = Vt + 0.125 V
⇒ 112.11 = 60 + 0.125 V
⇒ V = 8 x 52.11 = 417 m/sec
 
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