AAMC official guide #28 C/P & General Waves question

[sanguinee]

The explanation states that "propagation speed of a transverse wave (such as the sound wave in the vocal chords)" -- isn't this not true since sound is a longitudinal wave? Additionally, what is the rule of thumb for changes in frequency/velocity/wavelength for waves?

Another question also explains that when a wall is between a sound wave, it is reflected yet the sound on the other side of the wall is still the same frequency and wavelength as the unimpeded sound -- how is this possible?

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[aldol16]

1) Sound can be both transverse and longitudinal when traveling through a solid.

2) Sound energy relates to intensity, or amplitude of the wave. Part of the energy of the wave goes back (reflected) and part of it goes on (transmitted). The intensity of the sound is thus muffled on the other side of the wall but wavelength and frequency are unchanged. Perhaps you are thinking of the relationship E = h*f and questioning how such a reflection can obey conservation of energy. Your error there is that E = h*f gives the energy of a photon, not a sound wave.

 

[sanguinee]

1) Sound can be both transverse and longitudinal when traveling through a solid.

2) Sound energy relates to intensity, or amplitude of the wave. Part of the energy of the wave goes back (reflected) and part of it goes on (transmitted). The intensity of the sound is thus muffled on the other side of the wall but wavelength and frequency are unchanged. Perhaps you are thinking of the relationship E = h*f and questioning how such a reflection can obey conservation of energy. Your error there is that E = h*f gives the energy of a photon, not a sound wave.

When the sound is traveling through the wall, isn't the speed increase since it is through a solid, and this will result in a change in v and since v =f*wavelength, the frequency or wavelength would also change?

 

[aldol16]

When the sound is traveling through the wall, isn't the speed increase since it is through a solid, and this will result in a change in v and since v =f*wavelength, the frequency or wavelength would also change?

The explanation isn't making any assumptions about what happens within the wall. It's explaining what happens on the side the listener is on.

 

[wheatthinners]

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[aldol16]

Does this mean that a higher pitched sound (frequency) will sound louder than a lower pitched sound, even if they are at the same amplitude? And when velocity of a sound goes down (say, when going from solid -> gas, since bulk modulus will go down) does that decrease the intensity/loudness of a sound?

Not sure about the frequency. A change in material will change not only velocity but also density, so you have competing factors there.

 

[theonlytycrane]

for question #28, doesn't v = sqrt(tension / linear density)? I think choice B would also work

 

[sanguinee]

Not sure if this helps anyone else that has a similar question, but I found this on another website:

"frequency of a sound wave is defined as "the number of waves per second."
If you had a sound source emitting, say, 200 waves per second, and your ear (inside a different medium) received only 150 waves per second, the remaining waves 50 waves per second would have to pile up somewhere — presumably, at the interface between the two media.
After, say, a minute of playing the sound, there would already be 60 × 50 = 3,000 delayed waves piled up at the interface, waiting for their turn to enter the new medium. If you stopped the sound at that point, it would still take 20 more seconds for all those piled-up waves to get into the new medium, at 150 waves per second. Thus, your ear, inside the different medium, would continue to hear the sound for 20 more seconds after it had already stopped.
Besides, imagine that you switched the media around: now the sound source would be emitting 150 waves per second, inside the "low-frequency" medium, and your ear would receive 200 waves per second inside the "high-frequency" medium. Where would the extra 50 waves per second come from? The future? Or would they just magically appear from nowhere?'

" Frequency, in physics, is the number of crests that pass a fixed point in the medium in unit time.
So it should depend on the source not on the medium. If I take a source who vibrates faster than yours then number of crests that my source can create per second (for example) will be more than yours.
But speed of the wave depend on the properties of the medium, for example temperature, density etc etc.
v=wavelength*f
From this equation wavelength depend on the speed of the wave (i.e the medium) and the frequency so it is different for different medium. (Think of frequency in the equation as a constant since its only depend on the source, so if now speed changes; i.e medium changes then only wavelength changes) "

 

[NextStepTutor_2]

The explanation states that "propagation speed of a transverse wave (such as the sound wave in the vocal chords)" -- isn't this not true since sound is a longitudinal wave? Additionally, what is the rule of thumb for changes in frequency/velocity/wavelength for waves?

View attachment 201846



Another question also explains that when a wall is between a sound wave, it is reflected yet the sound on the other side of the wall is still the same frequency and wavelength as the unimpeded sound -- how is this possible?

View attachment 201848

Yes most of the time we deal with it on the MCAT, sound can be considered a prototypical longitudinal wave. We often see sound transmitted through gases and liquids, where it propagates as a longitudinal wave/compression wave. Through solids (like vocal cord tissue) though, sound it can be transmitted as both longitudinal waves and transverse waves.

As for your shortcuts, changing mediums cannot change the frequency of the sound wave, but it can change the speed and wavelength. Only changing the sound at the source will effect the frequency emitted.

They are related through the equation velocity = (freq)(wavelength)

As for why you will hear muddled or faint sounds though a wall (but you can still recognize voices) is due to the fact that when sound hits a surface such as a wall or a floor, several things happen. A proportion of the sound is:

1. Reflected back into the room (energy is preserved).
2. Transmitted through the surface and escapes from the room (energy is lost).
3. Absorbed by the surface (energy is lost).

As far as I know, the frequency would stay the same, but the transmittance of sound through the wall will depends on the frequency of the sound emitted (why you can always hear bass though a wall but higher notes, lyrics not so much. There are two things at work here:

1) Scattering: The size of the particles in the walls etc. will determine what frequency they scatter. That is, if the size of the particles is larger than the wavelength, then those waves will be scattered. If on the other hand the particles are smaller than the wavelength then those waves will pass through without being scattered.

2) Resonance: The walls have their resonant frequencies and being large objects, these tend to be low frequencies. Therefore bass sound waves are better at triggering resonance in the walls, which helps their transmittance.

My educated guess is that they are predicting a wavelength or speed change as the sound enters the wall from air, but then the reverse occurs when the sounds goes from wall to air again on the other side. this, to me at least, would help explain why the wavelength is the same as was emitted by the source.

Hope this helps, good luck!