Struggling with Homework: Seeking Guidance on Equations

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SUMMARY

The discussion centers on solving a physics homework problem involving sound waves and the calculation of time based on distance and speed. Participants clarify the equations needed, specifically using the formula t = d/v to determine time and the relationship between distance and echo time. The final conclusion establishes that the difference in echo times from two walls at distances d and d+w is given by the formula 2w/v, leading to a frequency calculation of 230 Hz based on the derived equations.

PREREQUISITES
  • Understanding of basic physics concepts, particularly sound waves
  • Familiarity with the equation t = d/v for calculating time
  • Knowledge of frequency calculations and their relationship to wave properties
  • Ability to manipulate algebraic equations for problem-solving
NEXT STEPS
  • Study the principles of sound wave propagation and reflection
  • Learn about the speed of sound in various mediums
  • Explore advanced applications of the equation t = d/v in different contexts
  • Investigate the relationship between frequency, wavelength, and speed of sound
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Students studying physics, educators teaching sound wave concepts, and anyone interested in understanding the mathematical relationships in wave mechanics.

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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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