Sound wave of wavelength

  • Thread starter jai6638
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  • #1
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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 dont 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
 

Answers and Replies

  • #2
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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 dont 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?
 
  • #3
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(330) ( .50) = 165 m ...

165/ .70 = 235 waves?

damn. that doesnt sound right..
 
  • #4
Doc Al
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jai6638 said:
AnsA) V= (Lambda)(f)
f= V/Lambda = 330/.70= 471.4 Hz

Ans b) I dont 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].
 
  • #5
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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..
 
  • #6
Doc Al
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45,075
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You'd better get the same answer!
 
  • #7
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oh ok .. lol.. i thought the answer was wrong.. alright thanks a lot...finally understood it :)
 

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