Calculating Frequency and Distance of a Sound Wave

In summary, a sound wave with a wavelength of 0.70 m and velocity of 330 m/sec is emitted for 0.50 s. The frequency of the wave is 471.4 Hz, and in this time interval, 235 complete waves are emitted. After 0.50 s, the front of the wave is 165 m from the source of the sound.
  • #1
jai6638
263
0
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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  • #2
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?
 
  • #3
(330) ( .50) = 165 m ...

165/ .70 = 235 waves?

damn. that doesn't sound right..
 
  • #4
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].
 
  • #5
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
You'd better get the same answer!
 
  • #7
oh ok .. lol.. i thought the answer was wrong.. alright thanks a lot...finally understood it :)
 

FAQ: Calculating Frequency and Distance of a Sound Wave

What is a sound wave of wavelength?

A sound wave of wavelength is a type of mechanical wave that is characterized by its length or distance between two consecutive peaks or troughs. It is a disturbance that travels through a medium, such as air or water, and carries energy from one point to another.

How is the wavelength of a sound wave measured?

The wavelength of a sound wave is measured by the distance between two consecutive points of the same phase, such as two peaks or two troughs. It can also be measured by the distance between two points of maximum compression or rarefaction in the medium.

What is the relationship between wavelength and frequency in a sound wave?

The wavelength and frequency of a sound wave are inversely proportional to each other. This means that as the wavelength increases, the frequency decreases, and vice versa. This relationship is described by the equation: wavelength = speed of sound / frequency.

How does the wavelength of a sound wave affect its pitch?

The wavelength of a sound wave is directly related to its pitch. This means that as the wavelength increases, the pitch decreases, and vice versa. This is because a longer wavelength corresponds to a lower frequency and a lower frequency is perceived as a lower pitch.

Can the wavelength of a sound wave be changed?

Yes, the wavelength of a sound wave can be changed by altering the frequency or the speed of the wave. For example, in a medium with a fixed speed of sound, changing the frequency of the wave will result in a change in wavelength. Similarly, in a medium with a fixed frequency, changing the speed of the wave will also result in a change in wavelength.

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