Sound Waves: Help Solving Problems

[motherlovebone]

I have been stumped on these problems for about a half an hour now, and I need some big help on them!

Problem 1:
"A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. An echo is heard 2.60 s later. How far away is the reflecting object?"

I guessed that 2.60 s was the period, so I found the reciprocal to get the frequency. Once I did that, I put the speed of sound in for v in the equation v=frequency x wavelength. My answer, 891.8 m, sounds preposterous however.

Problem 2:
"The notes produced by a violin range in frequency from approximately 196 Hz to 2637 Hz. Find the possible range of wavelengths produced by the instrument when the speed of sound is 340 m/s."

For this one, would I use v=frequency x wavelength? I did 340 divided by 196, which was 1.735, then 340 divided by 2637, which was 0.129. So would the range of wavelengths be 0.129 to 1.735?

 

[LeonhardEuler]

I have been stumped on these problems for about a half an hour now, and I need some big help on them!

Problem 1:
"A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. An echo is heard 2.60 s later. How far away is the reflecting object?"

You know the distance, "d" traveled by the wave is equal to the velocity times the time: [tex]d=vt[/tex]. So if the sound is heard 2.6s later then the distance it must have traveled would be the distance from the boat to the wall plus the distance from the wall to the boat. In other words, twilce the distance fom the wall.

Problem 2:
"The notes produced by a violin range in frequency from approximately 196 Hz to 2637 Hz. Find the possible range of wavelengths produced by the instrument when the speed of sound is 340 m/s."

I don't know where to start, these were not explained in class today. Please help!

The wavelength, [itex]\lambda[/itex] is equal to to the speed of the wave, v, over the frequency, f:
[tex]\lambda=\frac{v}{f}[/tex]
So the maximum and minimum frequency give you the minimum and maxximum wavelength. By the way, in case your interested in where this formula came from, it makes sense. A frequency of, say, 2 Hz means that 2 waves pass you in 1s. If the waves are traveling at 10 meters every second and two pass you in a second, then they each must be 5 meters long, so [itex]\lambda=\frac{v}{f}[/itex]

 

[kyle8921]

I understand that solving problems involving sound waves can be tricky and require a deep understanding of the concepts involved. It is important to approach these problems systematically and use the proper equations to find accurate solutions.

For Problem 1, your approach was correct in using the equation v = frequency x wavelength. However, the period of 2.60 s is not the same as the frequency. The frequency is the number of cycles per second, while the period is the time it takes for one complete cycle. To find the frequency, you can take the reciprocal of the period, which in this case would be 1/2.60 s = 0.3846 Hz. Plugging this into the equation, we get 343 m/s = 0.3846 Hz x wavelength. Solving for the wavelength gives us 892 m, which is close to your initial answer but more accurate.

For Problem 2, your approach was also correct in using the equation v = frequency x wavelength. However, instead of dividing the speed of sound by the frequency, you should divide it by the lowest and highest frequencies to get the range of wavelengths. So the range of wavelengths would be 340 m/s / 196 Hz = 1.735 m and 340 m/s / 2637 Hz = 0.129 m.

I hope this helps you in solving these problems. Remember to always double check your units and use the correct equations for a successful solution. Don't hesitate to seek help from a teacher or fellow scientist if you are still struggling. Keep up the good work!