Will sound be affected by gravity in this simple problem?

  • Thread starter Thread starter iJamJL
  • Start date Start date
  • Tags Tags
    Gravity Sound
Click For Summary
SUMMARY

The discussion centers on calculating the time it takes for an echo to return from a motion detector located 2.61 meters above the lab floor, given the speed of sound in air at 337.8 m/s and an air temperature of 10.8°C. The correct approach involves using the formula for distance, s = v*t, leading to a calculated time of approximately 0.0154 seconds for the sound to travel down and back up. The initial confusion arose from miscalculating the time using the wrong formula, but the correct method confirmed the expected result.

PREREQUISITES
  • Understanding of basic physics concepts, specifically sound propagation.
  • Familiarity with kinematic equations, particularly s = v*t.
  • Knowledge of how to convert temperature to speed of sound in air.
  • Ability to perform basic algebraic manipulations to solve equations.
NEXT STEPS
  • Study the effects of temperature on the speed of sound in different mediums.
  • Learn about the principles of echo location and its applications in various fields.
  • Explore advanced kinematic equations and their applications in real-world scenarios.
  • Investigate the impact of atmospheric pressure on sound speed and propagation.
USEFUL FOR

Students in physics, educators teaching sound dynamics, and anyone interested in the practical applications of sound propagation in various environments.

iJamJL
Messages
58
Reaction score
0

Homework Statement


Assume the air temperature is 10.8o C, and the speed of sound in air is 337.8 m/s. Suppose the lab floor is 2.61 m beneath the motion detector. If the detector produces a click, how long would it take before the echo returned?




Homework Equations


s=Vi*t + (.5)*(a)*(t^2)


The Attempt at a Solution


I just wanted to make sure I'm doing this correctly.

When going down:
distance = Vi(t) + (.5)*a(t^2)
2.61*2 (double distance) = 0 + [9.81*(t^2)] / 2
10.44 / 9.81 = t^2
t1 = 1.0316
 
Physics news on Phys.org
For one, you need to give us a better idea of the problem as a lot of this seems vague.

Two, does that seem at all reasonable? Should sound take a whole second to travel only 2.6 meters?
 
Found my mistake. I actually solved the answer correctly the first time (distance = v*t), but I accidentally input my answer incorrectly online. Woops :rolleyes:

This thread can be deleted.
 

Similar threads

Replies
3
Views
4K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
23
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 17 ·
Replies
17
Views
3K