Calculating Frequency and Distance of a Sound Wave

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Homework Help Overview

The discussion revolves around calculating properties of a sound wave, specifically its frequency, the number of complete waves emitted over a time interval, and the distance traveled by the wave front. The subject area includes wave mechanics and sound properties.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between wave velocity, wavelength, and frequency. Questions arise regarding the calculation of the number of waves emitted in a given time and the interpretation of frequency in this context.

Discussion Status

Some participants have provided calculations for frequency and distance traveled, while others express uncertainty about the number of waves emitted. There is an ongoing exploration of how to relate frequency to the number of waves over time, with some guidance offered on using the formula f x t.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a noted confusion regarding the interpretation of frequency and its application to the problem at hand.

jai6638
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A sound wave of wavelength 0.70 m and velocity 330 m/sec is produced for .50 s.

A) What is the frequency of the wave?
B) how many complete waves are emitted in this time interval?
c) after .50 s, how far is the front of the wave from the source of the sound?

AnsA) V= (Lambda)(f)
f= V/Lambda = 330/.70= 471.4 Hz

Ans b) I don't know how to do this.. is this the frequency?

Ans c) Lambda = (.50) ( 330) = 165 m

Are the above answers correct? also, i'd appreciate it if you guys could help me for the second question ( part b )

thanks
 
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jai6638 said:
A sound wave of wavelength 0.70 m and velocity 330 m/sec is produced for .50 s.

A) What is the frequency of the wave?
B) how many complete waves are emitted in this time interval?
c) after .50 s, how far is the front of the wave from the source of the sound?

AnsA) V= (Lambda)(f)
f= V/Lambda = 330/.70= 471.4 Hz

Ans b) I don't know how to do this.. is this the frequency?

Ans c) Lambda = (.50) ( 330) = 165 m

Are the above answers correct? also, i'd appreciate it if you guys could help me for the second question ( part b )

thanks
Seems correct, for question B think of it like this. If you emit the wave for a time so that it passes 1.4m how many complete waves will you have emitted?
Then if you emit for a time so that it passes 1.5m? And if it passes lambda*v?
 
(330) ( .50) = 165 m ...

165/ .70 = 235 waves?

damn. that doesn't sound right..
 
jai6638 said:
AnsA) V= (Lambda)(f)
f= V/Lambda = 330/.70= 471.4 Hz

Ans b) I don't know how to do this.. is this the frequency?
Think of it this way. The frequency (which you found in part a) is the number of cycles (or waves) produced each second. So... the number of waves produced in a given time must equal [itex]f \times t[/itex].
 
471.4 waves produced each second.

therefore, in half second = 471.4 * .5 ( f x t) = 235 waves?

i still get the same answer... doh..
 
You'd better get the same answer!
 
oh ok .. lol.. i thought the answer was wrong.. alright thanks a lot...finally understood it :)
 

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