Solving the Sound Wave Mystery: A Stone in a Well

[ace214]

[SOLVED] Sound wave

A stone is dropped from rest into a well. The sound of the splash is heard exactly 1.70 s later. Find the depth of the well if the air temperature is 12.0°C.


Anything _s is the sound wave, anything _f is the stone falling
I tried using these equations:


v_s = 331(sqrt(285/273) = 338.2

t_s = x/338.2
1.7 = t_s + t_f
x = .5(9.8)t_f^2

I plugged the first equation into the second, solved the second for t_f and plugged that into the last one. Ends up being a nasty quadratic but I'm pretty sure my math is ok. I ended up with

.000042841x^2 - .0492x + 14.161

but that gives imaginary solutions. Thanks for the help.


EDIT: Nevermind- I forgot to move the x over. Sorry!

 

[anathema]

Hello,

I would like to offer some clarification on the equations and calculations you have used in your attempt to solve this problem.

Firstly, the equation v_s = 331(sqrt(285/273) is used to calculate the speed of sound in air at a given temperature. In this case, the temperature is 12.0°C, which is equivalent to 285 Kelvin. However, this equation only applies to sound waves traveling horizontally through still air. In the case of a stone falling into a well, the sound wave is traveling vertically through air that is moving due to the stone's motion. Therefore, the speed of sound in this scenario would be slightly different and would need to be calculated using a different equation.

Secondly, the equation x = .5(9.8)t_f^2 is the correct equation to use for calculating the distance traveled by the stone (x) in a given time (t_f) when dropped from rest. However, in this case, we are not given the time it takes for the stone to hit the water, only the time it takes for the sound of the splash to reach the surface. This means that we cannot use this equation to solve for the depth of the well.

Instead, we can use the equation d = v_st_s, where d is the depth of the well, v_s is the speed of sound, and t_s is the time it takes for the sound to reach the surface. Rearranging this equation, we get d = v_st_s/2, since the sound wave has to travel down to the surface and then back up to the top of the well.

Plugging in the values given, we get d = (331(sqrt(285/273))(1.70)/2 = 143.7 meters.

I hope this helps clarify the problem and the correct approach to solving it. If you have any further questions, please don't hesitate to ask. Keep up the good work in your scientific pursuits!

 

[ogb p]

I would like to congratulate you on your attempt to solve the sound wave mystery. Your use of equations and calculations shows a strong understanding of the principles of sound and motion. However, I would like to offer some suggestions to improve your solution.

Firstly, it is important to note that the equation v_s = 331(sqrt(285/273) = 338.2 is an approximation and may not be accurate in all cases. It is based on the speed of sound in air at 0°C and may vary slightly at different temperatures. It would be more accurate to use the formula v_s = 331 + 0.6T, where T is the air temperature in °C.

Secondly, in the equation 1.7 = t_s + t_f, it is important to consider the time taken for the sound to travel from the stone to the bottom of the well and back to the surface. This should be twice the time t_s, as the sound wave has to travel twice the depth of the well. So the correct equation would be 1.7 = 2t_s + t_f.

Lastly, when solving the quadratic equation, it is important to consider the physical constraints of the problem. The depth of the well cannot be negative, so the negative solution should be discarded.

Overall, your approach to solving the sound wave mystery is commendable. However, it is important to carefully consider all the variables and constraints in order to arrive at an accurate solution. Keep up the good work!