The Logic Off Topic list

At 06:31 PM 9/10/01 -0400, you wrote:
>OUTPUT FREQ. FOR EACH tone generator and the resultant wavelength of each 
>sound at room temp.
>
>A= 10ms
>b= 0.1ms
>


Debbie, have you got access to a reference book on acoustics? If so, you
might get more out of this exercise if you can get clear on some of the
principles involved and try to nut things out yourself from first principles. 

Anyway ... 

>I also have to explain in as much detail what will happen when one of the 
>sound wave strikes the wall.

If a sound wave strikes a wall some sound will be transmitted. It will go
through the wall and be audible on the other side. Some will be reflected -
it will bounce right back. Some will be absorbed and just heat up the
absorbent material a little.

>Comment on the speed of the sound wave as it passes through the wall and the 
>resultant wavelength in general terms i. e without calculations

Sound travels faster in dense media -- liquids and solids -- than it does
in air. The period of the wave stays the same so the wavelength increases
in the dense medium. ( With light waves it is the other way -- it travels
slower in water or glass and hence the wavelength decreases.)

>Using an SPL meter (?) the sound level in the adjacent rooms is monitored and 
>readings taken for the output of each tone generator individually. Predict in 
>general terms the relationship between the readings for each freq. giving 
>reason for your predictions with reference to acoustic theory.

In order to absorb sound waves the thickness of the absorbtion layer should
ideally be 1/4 wavelength or larger.  So thin layers absorb high
frequencies and reflect low frequencies. To absorb low frequencies (<100Hz)
you need very thick layers of absorbing material.

>
>If the wall were covered in carpet the reflected energy of which freq. (A or 
>B) would be reduced most and why?

Carpet is typically 0.01 - 0.02 m thick.  10 kHz sound has a wavelength of
.034m. So it is easy to see why 1 or 2cm thick carpet which is about a
quarter wavelength for a 4 to 8 kHz sound wave damps out high frequencies
in the reflected and transmitted sound. 

One quarter wavelength for 100Hz is 3.4 Meters -- 11 feet or so. So carpet
makes little or no impact on reflection or transmission of 100Hz sound waves. 

>If there were a hole in the wall show using diagram (AHHHH) how the 
>propagation of the wave into the adjacent room would be different for two 
>freq.

Waves (sound or light) passing through an aperture (hole)  diffract --
which is to say they disperse or spread out -- and the longer the
wavelength the more they bend from their original . The result of this is
that bass frequencies become omnidirectional after leaving a port in a
speaker box or passing through a hole in the wall while treble frequencies
tend to beam and at 10kHz they are quite directional.

>A microphone is positioned in the test area and it's output fed to the 
>oscilloscope.  The freq. of tone generator B is reduced to 110Hz and the 
>output from both tone generators combined resulting in 'beating'. Produce a 
>rough sketch of the input to the speaker viewed via the oscilloscope, 
>describe the resultant aural effect and explain the acoustic theory behind 
>this phenomena from first principles.

Do you understand  the idea of superposition of waves? When waves are added
together the resultant wave can be found by simply adding the amplitudes
(height of the curve) of each wave at points along the curves of the
graphed waves. What you find is that waves reinforce one another at times
and cancel each other at other times.

For examples:

A) two identical audio waves, half a cycle_ out_ of phase cancel one
another. Result = silence.

B) Two identical waves_ in_ phase result in a wave with twice the amplitude.

C) If there is a frequency difference between two waves which are being
added, then they will reinforce one another and then cancel one another
cyclically. This cycle repeats at a frequency = F1 minus F2 where F1 an F2
are the freqencies of the two waves being added. 

So in the case of two sound waves -- one at 100 Hz and another at 110 Hz --
being added together the cycle of reinforcment and cancellation will have a
frequecy of 110 -110 = 10 Hz and will be perceived as and audible beat of
10 Hz.

Regards,
Murray