Struggling with Homework: Seeking Guidance on Equations

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Homework Help Overview

The discussion revolves around understanding the application of equations related to sound and distance, particularly in the context of calculating the time it takes for an echo to return after a clap. Participants explore the relationship between distance, speed, and time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants express uncertainty about how to apply relevant equations to the problem. Some suggest visualizing the scenario with a clap and echo to facilitate understanding. Questions arise regarding the implications of distance on time and the differences in time taken for echoes from walls at varying distances.

Discussion Status

The discussion has progressed with participants offering guidance on the relationships between distance, speed, and time. Some have provided specific equations and calculations, while others are still seeking clarity on the concepts and implications of their findings.

Contextual Notes

There are indications of confusion regarding the application of formulas and the interpretation of time differences in the context of sound travel. Some participants express a lack of initial attempts at solutions, highlighting the challenge of starting the problem.

OmniNewton
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Homework Statement



54a9cd5f702fff5fd1b3f88622813723.png

Homework Equations


Refer to 3.

The Attempt at a Solution


My problem is the theory I am unsure how any equations apply to this problem. This seems bad that I have no attempt at solution but I am honestly unsure how to start or of any relevant equations.
 
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OmniNewton said:

Homework Statement



54a9cd5f702fff5fd1b3f88622813723.png

Homework Equations


Refer to 3.

The Attempt at a Solution


My problem is the theory I am unsure how any equations apply to this problem. This seems bad that I have no attempt at solution but I am honestly unsure how to start or of any relevant equations.
Think instead of looking after equations. Imagine, you stand at distance d from a wall, and you clap once. The sound reflects from the wall. What time after the clap does the echo reach you?
 
ehild said:
Think instead of looking after equations. Imagine, you stand at distance d from a wall, and you clap once. The sound reflects from the wall. What time after the clap does the echo reach you?

v=dt
so t = v/2d
 
OmniNewton said:
v=dt
so t = v/2d
You mean the time is the shorter the longer is the distance ?
What is the correct answer?
And what is the time if the wall is at distance d+w?
What is the difference of times if one wall is at distance d and the other wall is w distance farther?
 
ehild said:
You mean the time is the shorter the longer is the distance ?
What is the correct answer?
And what is the time if the wall is at distance d+w?
What is the difference of times if one wall is at distance d and the other wall is w distance farther?

Sorry about the delay I've been really thinking this through. I'm just unsure what is being said by difference of times.
 
The first of your problem is how do you get the time you need to cover a distance d with speed v. You travel with speed 100 km/h, what time is needed to reach a town at 200 km distance?
 
ehild said:
The first of your problem is how do you get the time you need to cover a distance d with speed v. You travel with speed 100 km/h, what time is needed to reach a town at 200 km distance?
2 hours
 
What is the general formula used to get time from distance and speed?
 
ehild said:
What is the general formula used to get time from distance and speed?
ehild said:
What is the general formula used to get time from distance and speed?
d=vt
t = d/v
 
  • #10
OmniNewton said:
d=vt
t = d/v
It is correct now. In what time does the echo of the clap return from a wall at distance d? And what is the time of return from a wall at distance d+w?
 
  • #11
The time taken to return back(when the distance is d)
is equal to 2d/v where v is the velocity of sound
and when the distance is d+w
The time taken is 2(d+w)/v
Therefore
The difference in the times taken equals 2w/v
 
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  • #12
Alright after looking at Uchiha's work that is correct. difference in time represents the period.

So,

f = 1/T = v/2w = 343 m/s/2(0.75m) = 230 Hz

for b) since f is proportional to 1 /w the frequency will be higher
 
  • #13
Precisely,OmniNewton
Just some thinking and calculations always suffice :)
 
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  • #14
Thank You So Much!
 
  • #15
You're welcome!
 
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  • #16
Heh thanks mate for those likes!:)
 

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