Drop a stone in a well - Waves problem

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SUMMARY

The discussion focuses on calculating the total time it takes to hear a splash after dropping a stone into a 7.35 m deep well. The time for the stone to fall is calculated using the equation \( t_{1} = \sqrt{\frac{2d}{g}} \), yielding 1.22 seconds. The time for the sound to travel back up the well is calculated as \( t_{2} = \frac{4d}{v} \), resulting in 0.0857 seconds. The total time calculated was 1.3 seconds, which is incorrect; the correct answer is 1.24 seconds. The discrepancy arises from misapplying the open-closed tube model to the well scenario.

PREREQUISITES
  • Understanding of kinematic equations, specifically \( d = \frac{1}{2}gt_{1}^{2} \)
  • Knowledge of sound wave propagation in air, including speed of sound (343 m/s)
  • Familiarity with wave properties, including wavelength and frequency relationships
  • Basic physics concepts regarding free fall and sound travel time
NEXT STEPS
  • Review the principles of free fall and gravitational acceleration
  • Study sound wave propagation in different mediums, focusing on air
  • Learn about the differences between open-closed and closed-closed tube models in acoustics
  • Explore the implications of depth and medium on wave behavior in physics problems
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Students studying physics, particularly those focusing on kinematics and wave mechanics, as well as educators looking for examples of sound propagation in real-world scenarios.

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Homework Statement


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)

Homework Equations


d=\dfrac{1}{2}gt_{1}^{2}

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

The Attempt at a Solution


The time taken for the stone to drop in water:
t_{1}=\sqrt{\dfrac{2d}{g}}=1.22 s

The time taken for the reflected wave to make it outside the well
v=\dfrac{4d}{T}
So, T=\dfrac{4d}{v}=\dfrac{4(7.35)}{343}=0.0857 s

Finally,
t_{total}=t_{1}+t_{2} = 1.22 s + 0.0857 s = 1.3 s

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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You have the time for the stone to fall and hit the water correct = 1.22s
This makes a noise and the sound has to travel to the top of the well, 7.35m and the speed of sound = 343m/s. Calculate the time taken for sound to travel this distance.
How much time has passed since you dropped the stone??
 
But what about the wavelength of the sound wave? I can't change the wavelength.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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