# Wavelength problem

by cytochrome
Tags: wavelength
 P: 162 1. The problem statement, all variables and given/known data 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. 2. Relevant equations speed of sound = 350 m/s λ = speed/frequency 3. 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?
Emeritus
PF Gold
P: 5,196
 Quote by cytochrome 1. The problem statement, all variables and given/known data 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. 2. Relevant equations speed of sound = 350 m/s λ = speed/frequency 3. 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?
 Emeritus Sci Advisor PF Gold P: 5,196 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...
P: 162
Wavelength problem

 Quote by cepheid 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?
Emeritus
PF Gold
P: 5,196
 Quote by cytochrome 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...
 Emeritus Sci Advisor PF Gold P: 5,196 Now I think I get it. You have to be careful when talking about longitudinal waves to distinguish between talking about the displacment 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 displacment 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 displacment 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:  | | | | |||| and this  |||| | | | | 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.
P: 162
 Quote by cepheid Now I think I get it. You have to be careful when talking about longitudinal waves to distinguish between talking about the displacment 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 displacment 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 displacment 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:  | | | | |||| and this  |||| | | | | 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
Emeritus