Reducing Vibration in Home: How & Why?

[Sarahborg]

Homework Statement

How can one reduce the vibrations entering the house? And why do certain sounds vibrate more in certain place, but quiet fown if you move a bit further away.

Homework Equations

The Attempt at a Solution

I was thinking it's something to do with the fact that sounds wave are mechanical waves, they need a medium to be propagated. I thought maybe a vacuum or something of the sort could help.

 

[NFuller]

How can one reduce the vibrations entering the house? And why do certain sounds vibrate more in certain place, but quiet fown if you move a bit further away.

Is there a diagram for this problem? The problem as stated is far to vague for us to give anything but wild guesses.

 

[Sarahborg]

No, there isn't any diagram. It is just a theory question where we are meant to explain how vibrating sounds (ex those from air-conditioners) entering the house could be reduced.

Thanks

 

[haruspex]

maybe a vacuum or something of the sort

Rather impractical.
Think about the paths the sound takes to get into the house. What do the vibrations have to pass through?

For the next part, what do you know about standing waves?

 

[epenguin]

Sound in air or other gas is a pressure wave transmitted essentially by molecules colliding. Vacuum - no molecules - no sound transmitted. I think this was first demonstrated by Robert Boyle who rigged up a bell he could shake and then created a vacuum, and no sound came out any more. He was a pioneer in creating vacuums, hi-tech for the epoch, and a lot of our concepts, almost unconscious are due to him and a few others of his time clarifying how air and gases were real substances. His experiments established things now so obvious that nobody ever thinks about or mentions them - although sound was not transmitted he showed that electrical and magnetic forces and of course light were. (subject to correction this is my memory of things read long ago.)

 

[haruspex]

Sound in air or other gas is a pressure wave transmitted essentially by molecules colliding. Vacuum - no molecules - no sound transmitted.

True, but as I posted, that would be rather an impractical way to stop noise entering a house. There are easier ways.

 

[Sarahborg]

Vibrations pass through air.

We only dealt with mechanical waves. We just mentioned that stationery waves are found for example in string instruments.

 

[Sarahborg]

True, but as I posted, that would be rather an impractical way to stop noise entering a house. There are easier ways.

I was initially thinking of double glazing or cavity walls. However, these are quite impractical as well since one assumes the house is already built.

 

[haruspex]

Vibrations pass through air.

What if all the doors and windows are closed?

double glazing

Yes. One of the reasons for adding that to a house is to reduce incoming noise.

We just mentioned that stationery waves are found for example in string instruments.

Ok, but stationary waves are key to answering the second question.

 

[Sarahborg]

So apart from closing all doors and windows (which is not that ideal in summer) and double glazing, what other options would you suggest?

Thank you very much for your patience!

 

[haruspex]

So apart from closing all doors and windows (which is not that ideal in summer) and double glazing, what other options would you suggest?

Thank you very much for your patience!

No, I'm not suggesting closing all doors and windows. I am pointing out that sound still gets through even if they were all closed.
Even in summer, it would be unusual, if not impossible, to have all doors and windows open. Only a portion of windows can be opened, generally.
So, how does sound get through a closed door or window, and what might you be able to do to reduce it?

 

[Sarahborg]

So if you close the door or a window, air isn't entering the room you're in, but sound waves travel through the glass of the window or the wood of the door.

The problem is how to stop sound from entering. Maybe you can use soundproof materials that absorb sound?

 

[haruspex]

So if you close the door or a window, air isn't entering the room you're in, but sound waves travel through the glass of the window or the wood of the door.

The problem is how to stop sound from entering. Maybe you can use soundproof materials that absorb sound?

Yes, that would certainly help stop sound coming through the walls. In what other way would that help?
Windows are more of a problem because sound absorbing materials might tend to be light absorbing too. We've mentioned double glazing. (What do you think is the reason that helps?)

 

[Sarahborg]

Yes, that would certainly help stop sound coming through the walls. In what other way would that help?

Maybe the vibrations are less powerful? I'm not sure on this one.

Windows are more of a problem because sound absorbing materials might tend to be light absorbing too. We've mentioned double glazing. (What do you think is the reason that helps?)

Double glazing works because of the thin air gap layer between the two glass panes, so sound doesn't enter, since some of it is reflected back

 

[haruspex]

Maybe the vibrations are less powerful?

Think about what happens to the sound after it has come into a room.

Double glazing works because of the thin air gap layer between the two glass panes, so sound doesn't enter, since some

Yes. Each transition to a different medium results in some reflections, increasing the opportunity for absorption.

 

[Sarahborg]

It echoes depending on how much the room is full or empty (as in it echoes less in a room with a lot of furniture)

 

[haruspex]

It echoes depending on how much the room is full or empty (as in it echoes less in a room with a lot of furniture)

Yes, and any soundproofing on the walls will help deaden reflections too.

Now to the second question, why does the sound intensity vary from place to place within a room?

 

[Sarahborg]

Maybe it varies depending on how close or far away one is from the source of the sound?

 

[haruspex]

Maybe it varies depending on how close or far away one is from the source of the sound?

No. As I wrote, it is to do with standing waves, though it perhaps would have been more helpful to say interference patterns.

 

[Sarahborg]

No. As I wrote, it is to do with standing waves, though it perhaps would have been more helpful to say interference patterns.

What does it have to do with standing waves and interference patterns? And what are interference patterns exactly? I am a bit lost here.

Also, how would you detect the source of a sound if it is not heard the same within a room and how close or far away you are does not make any difference? And to reduce the vibrations, would there be a simpler way than double glazing or soundproof materials?

Thanks and sorry for asking too many questions!

 

[haruspex]

what are interference patterns exactly?

Google that and standing waves. Ask again here if you have further questions.

how would you detect the source of a sound if it is not heard the same within a room and how close or far away you are does not make any difference?

It can be hard to tell where a sound is coming from when inside a house. If the source is at the front of the house but the only windows open are at the back you can come to the wrong belief.

to reduce the vibrations, would there be a simpler way than double glazing or soundproof materials?

Not that I am aware of.

 

[Sarahborg]

I did, but still didn't understand how sound vibrations vary within the same room.

 

[haruspex]

I did, but still didn't understand how sound vibrations vary within the same room.

You will have seen diagrams of resonance in a half open tube. At the closed end, the air cannot move, so a node occurs.
The same happens at the wall of a room.
With x as the distance from the wall, and a suitable choice of time 0, the incident sound wave is A sin(αx+ωt), where the wavelength λ= 2π/α and the speed is ω/α. The reflected wave goes at the same speed and frequency in the opposite direction, so it must have the form A'sin(αx-ωt+φ), for some constant φ. Ignoring energy loss, A=A'.
Since there is a node at the wall we have a zero sum there:
A sin(-ωt)+A sin(ωt+φ)=0 for all t, so φ=0.
At x from the wall, the sum gives A(sin(αx+ωt)+sin(αx-ωt)).
Applying the formula sin(A+B)=sin(A)cos(B)+cos(A)sin(B) we get 2Asin(αx)cos(ωt).
Ths is a standing wave. At a given value of x, 2Asin(αx) is constant, so just behaves as the amplitude of the visible oscillation cos(ωt).
In particular, at x= nλ/2, the amplitude is 2Asin(nπ)=0. So there is a node at each multiple of a half wavelength from the wall. In principle, if you stand at a node you will hear nothing; if you stand at an antinode you will hear amplitude 2A.
Of course, any real world sound will be a mix of frequencies, with different frequencies forming nodes at different distances from the wall.

I looked on the net for an animation of this. Plenty of still images, but didn't find any good animations.

 

[Sarahborg]

I looked on the net for an animation of this. Plenty of still images, but didn't find any good animations.

I found this

I really can't thank you enough for your very detailed explanation. I got the gist of it. Thanks!