Fundamental frequency of a standing wave

In summary, the homework statement states that a standing wave is established in a string of length 150 cm, and that the string vibrates in four segments when driven at 140 Hz. The wavelength in meters is found to be 3m. The fundamental frequency is found to be 35 Hz.
  • #1
dlthompson81
44
0

Homework Statement



A standing wave is established in a string of length 150 cm fixed at both ends. The string vibrates in four segments when driven at 140 Hz.

Find the wavelength in meters.

Find the fundamental frequency.

Homework Equations



L = Nλ
v = fλ

The Attempt at a Solution



L = Nλ
1.5m = 2λ λ = 0.75m (This answer was correct)


v = fλ
v = 140(0.75) = 105
L = Nλ
1.5m = (1/2)λ λ = 3
v = fλ
105=f3 f = 35 Hz (This was wrong)

Where did I go wrong on this? Thanks for the help.
 
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  • #2
dlthompson81 said:

Homework Statement



A standing wave is established in a string of length 150 cm fixed at both ends. The string vibrates in four segments when driven at 140 Hz.

Find the wavelength in meters.

Find the fundamental frequency.

Homework Equations



L = Nλ
v = fλ

The Attempt at a Solution



L = Nλ
1.5m = 2λ λ = 0.75m (This answer was correct)


v = fλ
v = 140(0.75) = 105
L = Nλ
1.5m = (1/2)λ λ = 3
v = fλ
105=f3 f = 35 Hz (This was wrong)

Where did I go wrong on this? Thanks for the help.

fo should be related to f by the number of nodes
The only other explanation is a f0 of 70hz that is referring to the segments of the standing wave + original wave in counting segments
 
  • #3
If the string vibrates with only one segment, what is the wavelength?
 
  • #4
If only one segment is vibrating, then the wavelength is 2 * Length. Since the string is 1.5m long, that would make the wavelength 3m. Right?
 
  • #5
dlthompson81 said:
If only one segment is vibrating, then the wavelength is 2 * Length. Since the string is 1.5m long, that would make the wavelength 3m. Right?
Right.
 
  • #6
To get the frequency I need to divide the speed of the wave by the wavelength.

The speed is the frenquency * wavelength.

So v = 140Hz * 3 = 420

Then v = fλ:
420 = f3
f = 140 ?

Is that it?
 
Last edited:
  • #7
Ok. So I tried that answer. It's not right either. I'm totally lost I guess.
 
  • #8
The speed is the same for all the waves, and f·λ = speed of the wave, so ...
 
  • #9
The speed is 420.

I'm not sure of what wavelength to use.

3 was the wavelength at 2*L

0.75 was the wavelength for the first part of the problem when there were 4 segments on the string

420/.75 = 560 doesn't sound right, if the wavelength gets longer and speed stays constant, then f should be getting smaller

420/3 was wrong, 3 was the wavelength if there was only one section on the string which i thought was the correct wavelength
 
  • #10
3 is also the wavelength of the fundamental.
 
  • #11
So, wouldn't it be:

v = f λ

420 = f 3

420/3 = 140

Or, did I mess up on the speed somehow?
 
  • #12
Didn't you calculate the speed as 105 m/s ?

OK, that gives 35 Hz, which you said was wrong.
 
  • #13
Yes. But I thought I had it wrong.

I originally had the problem in the original post as this:

105=f3 105/3 f=35

Which was wrong according to my homework program.
 
  • #14
I found several other example problems that I worked out the same exact way and got those right, but when I worked it out for my homework and entered it in, it was wrong. I don't know where I went wrong.
 
  • #15
Ok. So the due date was this morning, and I can download the answers now. I'm going to discuss it with my teacher because I don't believe the answer was correct.

The answer was:

f = v/2L = fλ/2L = (140 Hz)(0.75 m) / 2(150 cm) = 0.35 Hz

I'm thinking the m and cm is a mistake. I think it should all be in units of meters or centimeters.
 
  • #16
dlthompson81 said:
Ok. So the due date was this morning, and I can download the answers now. I'm going to discuss it with my teacher because I don't believe the answer was correct.

The answer was:

f = v/2L = fλ/2L = (140 Hz)(0.75 m) / 2(150 cm) = 0.35 Hz

I'm thinking the m and cm is a mistake. I think it should all be in units of meters or centimeters.

Are you sure the driving frequency wasn't 1.40 Hz, rather than 140 Hz. That would explain it also. 1.40 Hz sounds more reasonable for a string.
 
  • #17
I talked to the teacher about it. He said it was a mistake in the problem. He gave everyone credit for it.

Thanks for the help with it. I hate I took up your time with an unsolvable problem.
 

What is the definition of fundamental frequency of a standing wave?

The fundamental frequency of a standing wave is the lowest frequency at which a standing wave pattern can occur in a given system. It is also known as the first harmonic frequency.

How is the fundamental frequency of a standing wave calculated?

The fundamental frequency of a standing wave can be calculated by dividing the wave speed by twice the length of the standing wave. This can be expressed as f = v/2L, where f is the fundamental frequency, v is the wave speed, and L is the length of the standing wave.

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

The wavelength of a standing wave is equal to twice the length of the standing wave, which is directly related to the fundamental frequency. As the fundamental frequency increases, the wavelength decreases, and vice versa.

How does the tension and density of the medium affect the fundamental frequency of a standing wave?

The fundamental frequency of a standing wave is directly proportional to the tension in the medium and inversely proportional to the density of the medium. This means that an increase in tension will result in an increase in the fundamental frequency, while an increase in density will result in a decrease in the fundamental frequency.

What are some real-life examples of standing waves and their fundamental frequencies?

Standing waves can be found in musical instruments, such as strings and air columns, where the fundamental frequency determines the pitch of the sound produced. They can also be observed in vibrating structures, such as bridges and buildings, with their fundamental frequencies determining their natural resonant frequencies.

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