(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

You drop a stone from rest into a well that is 7.35 m deep. How long does it take before you

hear the splash?

d=7.35 m

v=343 m/s (because the disturbance is air molecules and the medium is air not water)

2. Relevant equations

[itex]d=\dfrac{1}{2}gt_{1}^{2}[/itex]

[itex]\lambda=4d[/itex] and [itex]v=\lambda f[/itex] and we know that f is [itex]f=\dfrac{1}{T}[/itex]

3. The attempt at a solution

The time taken for the stone to drop in water:

[itex]t_{1}=\sqrt{\dfrac{2d}{g}}=1.22 s[/itex]

The time taken for the reflected wave to make it outside the well

[itex]v=\dfrac{4d}{T}[/itex]

So, [itex]T=\dfrac{4d}{v}=\dfrac{4(7.35)}{343}=0.0857 s[/itex]

Finally,

[itex]t_{total}=t_{1}+t_{2} = 1.22 s + 0.0857 s = 1.3 s[/itex]

However this is wrong!! the answer is 1.24 seconds. What am i doing wrong?

My first initial thought is to think of the well as an open-closed tube. When those relevant equations are applied to a open-closed tube I get the wrong answer. Why is thinking of the well as a open-closed tube wrong ?

1. Is it because a little portion of the well at the bottom is filled with water so that the ENTIRE well is not actually 7.35 m ? OR

2. Is it because we have no information about its wavelength regardless if it's a open-closed tube? OR

3. Does the wavelength calculated above does not make sense ? (e.g. How can the wavelength be 29.4 m when the well itself is 7.35 m ? (Assuming the well as a long cylinder)

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# Homework Help: Drop a stone in a well - Waves problem

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