# Wavelength Problem: Compute Frequency in 4m Room

• cytochrome
In summary, the longest sound wave that can occur in a room that is 4m wide in the x direction (ignoring the y and z directions) is 8m with a frequency of 87.5 Hz. This is because the displacement of the air and the pressure of the air are out of phase, resulting in a slinky effect where the air oscillates between maximum and minimum pressure at the edges and centre of the room. This creates a wavelength of 2L, with L being the length of the room, and since the boundary conditions are fixed at the ends, the wavelength is halved to fit inside the room, resulting in a longest wavelength of 8m and a frequency of 87.5 Hz.
cytochrome

## Homework Statement

A room is 4m wide in the x direction, ignore the y and z directions. Compute the longest sound wave that can occur in the room and its frequency.

## Homework Equations

speed of sound = 350 m/s

λ = speed/frequency

## The Attempt at a Solution

Since the room is 4m wide and sound is a stationary wave, the longest wavelength that can occur is 4m.

λ = 4m

frequency = speed/λ = 350/4 = 87.5 Hz

This just seems incredibly simple, is there anything I'm missing?

cytochrome said:

## Homework Statement

A room is 4m wide in the x direction, ignore the y and z directions. Compute the longest sound wave that can occur in the room and its frequency.

## Homework Equations

speed of sound = 350 m/s

λ = speed/frequency

## The Attempt at a Solution

Since the room is 4m wide and sound is a stationary wave, the longest wavelength that can occur is 4m.

λ = 4m

frequency = speed/λ = 350/4 = 87.5 Hz

This just seems incredibly simple, is there anything I'm missing?

Yeah, there is. Consider by analogy standing waves (what you call stationary waves) on a string. The string is fixed at both ends. You can have a standing wave where the whole string just oscillates up and down. So there is just one node in the middle, and two anti-nodes at the ends. Or maybe it's the other way around, I forget the terminology. But the point is, there's one point in the middle with max transverse displacement, and two points at the ends with zero displacment. What is the wavelength of this wave, relative to the string length L?

Hmm, now I'm not so sure that I'm right. I answered hastily. For a longitudinal wave, if the centre of the room varies between max and min amplitude, this means there is alternately a compression and a rarefaction there (high pressure and low pressure). But for this to be true (compression in the centre), you'd think the air piling up there in the centre would result in areas of lower density and pressure on the sides. You can't just have the whole room oscillating between high and lower pressure, I don't think...

cepheid said:
Yeah, there is. Consider by analogy standing waves (what you call stationary waves) on a string. The string is fixed at both ends. You can have a standing wave where the whole string just oscillates up and down. So there is just one node in the middle, and two anti-nodes at the ends. Or maybe it's the other way around, I forget the terminology. But the point is, there's one point in the middle with max transverse displacement, and two points at the ends with zero displacment. What is the wavelength of this wave, relative to the string length L?

Thanks for that insight! That makes sense now... But I'm still confused about one thing

In the situation you mentioned, the wavelength is 2L/m, where m is an integer.

That's the wavelength when boundary conditions are involved, so what would m be? Would m =1 since we are talking about the longest wavelength and therefore the fundamental frequency?

How can a wavelength of 2L = 2*4 = 8 fit inside a 4 meter room?

cytochrome said:
Thanks for that insight! That makes sense now... But I'm still confused about one thing

In the situation you mentioned, the wavelength is 2L/m, where m is an integer.

That's the wavelength when boundary conditions are involved, so what would m be? Would m =1 since we are talking about the longest wavelength and therefore the fundamental frequency?

How can a wavelength of 2L = 2*4 = 8 fit inside a 4 meter room?

For the case of the string, when m = 1, you are right that λ = 2L = 8 m. If you think about it, when the whole string is vibrating up and down, this corresponds to one peak or one trough. So the answer is, a whole wavelength doesn't fit on the string in that scenario. The string is half a wavelength. It's like in this figure, right-most column, the top diagram of the three:

For the case of a sound wave, I'm still struggling with my intuition as to how that's possible. I would think that the rightmost column, middle diagram would be the shortest sound wave you could fit, since if the pressure and density are peaked in the middle, they must necessarily be lower on either side...

Now I think I get it. You have to be careful when talking about longitudinal waves to distinguish between talking about the displacement of the air and the pressure of the air, which are out of phase with each other. For a closed room (fixed boundary conditions) the edges are displacement nodes (because air there can't go anywhere), but pressure antinodes, (because the pressure at the ends varies between maximum and minimum as air coming in from the centre of the room piles up at the edges and then "unpiles"). Likewise, in the centre, there is a displacement antinode (because air moves the most there) but a pressure node (because it all moves by the same amount, so there is no pile up in the centre, unlike at the ends). So we have this slinky effect, oscillating between this:

Code:
|   |  |  | ||||

and this

Code:
|||| |  |  |   |

where, the closer the lines are together, the more compressed the air is, and the higher the density and pressure.

In the first figure, maximum rightward displacement of centre air corresponds to the upward "peak" in the m = 1 diagram above. In the second figure maximum leftward displacement of centre air corresponds to the downward "trough" in the m = 1 diagram above

The distance between nodes or antinodes is half a wavelength, and since we have nodes at the ends of the room (just like for the string with fixed ends) the answer is again λ/2 = L.

Last edited:
cepheid said:
Now I think I get it. You have to be careful when talking about longitudinal waves to distinguish between talking about the displacement of the air and the pressure of the air, which are out of phase with each other. For a closed room (fixed boundary conditions) the edges are displacement nodes (because air there can't go anywhere), but pressure antinodes, (because the pressure at the ends varies between maximum and minimum as air coming in from the centre of the room piles up at the edges and then "unpiles"). Likewise, in the centre, there is a displacement antinode (because air moves the most there) but a pressure node (because it all moves by the same amount, so there is no pile up in the centre, unlike at the ends). So we have this slinky effect, oscillating between this:

Code:
|   |  |  | ||||

and this

Code:
|||| |  |  |   |

where, the closer the lines are together, the more compressed the air is, and the higher the density and pressure.

The distance between nodes or antinodes is half a wavelength, and since we have nodes at the ends of the room (just like for the string with fixed ends) the answer is again λ/2 = L.

That was extremely intuitive, thanks! I was seriously thinking for hours about what it means to fit an 8m wave in a 4m room... but it's actually just air displacement. I guess I just always have the picture of an electromagnetic wave in my head

cytochrome said:
That was extremely intuitive, thanks! I was seriously thinking for hours about what it means to fit an 8m wave in a 4m room... but it's actually just air displacement. I guess I just always have the picture of an electromagnetic wave in my head

Again, like I said before for the string, all it means is that you only fit half a wave on the string, or in the room. (see attached)

There's no reason why you can't have an EM wave of wavelength λ = 2L in a resonant cavity of length L. It just means you have only half a wave in the cavity (again, see attached)

#### Attachments

• lambda=2L.png
1 KB · Views: 504

## 1. What is a wavelength problem?

A wavelength problem involves calculating the frequency of a wave given its wavelength and the speed of the wave, which is typically the speed of light for electromagnetic waves. This type of problem is commonly encountered in physics and related fields.

## 2. How do you compute frequency in a 4m room?

To compute frequency in a 4m room, you would need to know the speed of the wave, which is the speed of light in this case, and the wavelength of the wave. You can then use the formula f = c/λ, where f is the frequency, c is the speed of light, and λ is the wavelength. Plug in the values and solve for frequency.

## 3. Can you give an example of a wavelength problem?

One example of a wavelength problem is calculating the frequency of a radio wave with a wavelength of 5 meters. We know that the speed of light is 3x10^8 meters per second, so we can use the formula f = c/λ to find the frequency. The frequency would be 3x10^8 / 5 = 6x10^7 Hz.

## 4. How is frequency related to wavelength?

Frequency and wavelength are inversely related. This means that as the wavelength increases, the frequency decreases, and vice versa. This relationship is described by the formula f = c/λ, where c is the speed of light and λ is the wavelength.

## 5. What units are used to measure frequency and wavelength?

Frequency is typically measured in hertz (Hz), which is the number of cycles or waves per second. Wavelength is measured in meters (m) for electromagnetic waves, but can also be measured in other units such as nanometers (nm) or micrometers (μm) for different types of waves.

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