Light & Sound Waves: Thermal Energy & Frequency

[Pranav Jha]

1. As the sound wave propagates away from the source, the energy is "thermally eaten". So, there is a gradual decrease in kinetic energy of the vibrating particles. Doesn't that lower the frequency of the sound wave? (i first considered from the classical energy point of view and then wondered if E=hv could be applied here? Please answer using both notions)

 

[Rajini]

E=hv is only for photons or light.

 

[schip666!]

Being "thermally eaten" means the coherent sound pressure changes are dissipated into the medium -- usually air. Their kinetic energy is still there, it's just turned itself into heat or noise -- basically entropy increase.

 

[sophiecentaur]

The frequency must stay the same but the amplitude decays with distance.

 

[Pranav Jha]

The frequency must stay the same but the amplitude decays with distance.

Why is the frequency of a wave unaffected by its amplitude? Isn't it kind of counter-intuitive?

 

[Born2bwire]

Why is the frequency of a wave unaffected by its amplitude? Isn't it kind of counter-intuitive?

How so? Think of a simple harmonic oscillator like a spring. Regardless of how much you initially displace the spring from equilibrium, it will oscillate at the same frequency. Only the amplitude will change. We can abstractly think of a sound source as a harmonic driver to a damped mass-spring system. The source drives the oscillator at a driven frequency and then we cut off the source to show that the wave has moved away from the source. As the wave propagates, it loses energy and thus the oscillator is damped. In all of this, the frequency stays the same but the amplitude diminishes due to the damping.

 

[davenn]

Why is the frequency of a wave unaffected by its amplitude? Isn't it kind of counter-intuitive?

no, :) or following on from Born2bwire 's comments think of an audio oscillator or say a single tone coming out of a bit of sound gear you have. You can turn the volume (amplitude) up and down but the freq of that tone doesn't change.

Over a distance you loose the higher freqs in a given noise source due to attenuation
but the lower freq's still get through

Think of your neighbour's stereo up loud, you don't hear the singing or much of the high freq's of the lead guitar. what you hear is the low freq's of the bass guitar and the drums going thud thud

Dave

 

[sophiecentaur]

Why is the frequency of a wave unaffected by its amplitude? Isn't it kind of counter-intuitive?

You could ask your 'intuition' what would cause more distant air to vibrate at any frequency different from the vibrations of the nearer air. There will be a phase lag, of course and the amplitude will reduce as energy spreads out and is lost by friction effects. But adjacent molecules must be vibrating at the same frequency or else they would eventually get out of step and be vibrating in different directions. That really would be counter-intuitive.

 

[Pranav Jha]

i am not familiar with simple harmonic motion. What i was thinking was if the amplitude decreases, the displacement from the mean position decreases. So, the distance to be distance to be vibrated decreases and thus the decreased displacement should be covered quickly. So, the time period increases and frequency increases.
But it obviously is wrong! Please clear up my thinking in this regard

 

[compitentboy]

Can anyone of you solve me this question?
A Stationary wave is making 5 loops. The distance between two consecutive statiopnary points is 10cm and its velocity is 20m/s.What will be its frequency?What will be the fundamental harmonic frequency?
Thanks in anticipation.

 

[Pranav Jha]

I would want others to check me on this:

wavelength= 10*2= 20 cm=0.2m
velocity= 20m/s
frequency= velocity/wavelength=20/0.2=100 Hz

I think:
for fundamental harmonic frequency, there should be only one loop along the whole string
Length of the string= 5*10=50cm=0.5m
so: wavelength=0.5*2= 1 m
frequency= 20/1= 20 Hz

 

[russ_watters]

i am not familiar with simple harmonic motion.

That's the while issue. Sound waves are a form of simple harmonic motion. Springs, guitar strings, drums, your eardrums, and the air itself all follow the same basic rules and one of those rules is that amplitude and frequency are independent of each other.

 

[sophiecentaur]

i am not familiar with simple harmonic motion. What i was thinking was if the amplitude decreases, the displacement from the mean position decreases. So, the distance to be distance to be vibrated decreases and thus the decreased displacement should be covered quickly. So, the time period increases and frequency increases.
But it obviously is wrong! Please clear up my thinking in this regard

I can see your problem here. But you need to consider one more factor. The 'restoring' force from, say a spring on a suspended mass (a very simple example of SHM) is proportional to the amount of displacement. So when the oscillations are small, the forces are also small and when the oscillations are big then so are the restoring forces.The overall result is that the period of oscillation is the same for all amplitudes of oscillation.
There is a story of Galileo sitting in church and realising that a swinging lamp had a constant time period - he 'invented' / discovered the Pendulum, which has the same period whatever the size of the swing. (That is an ideal pendulum and, for the purists who will want to reply and say it's wrong!, it doesn't apply for large amplitudes).

 

[Rajini]

I would want others to check me on this:

wavelength= 10*2= 20 cm=0.2m
velocity= 20m/s
frequency= velocity/wavelength=20/0.2=100 Hz

I think:
for fundamental harmonic frequency, there should be only one loop along the whole string
Length of the string= 5*10=50cm=0.5m
so: wavelength=0.5*2= 1 m
frequency= 20/1= 20 Hz

I think length of the string is wrong ? Where you got that formula? Why you recalculate wavelength and frequency ? (seems like it is a given data which is 0.2 m)..

 

[Pranav Jha]

I think length of the string is wrong ? Where you got that formula? Why you recalculate wavelength and frequency ? (seems like it is a given data which is 0.2 m)..

what do you mean? i don't think they are given

 

[Rajini]

Please read your post properly.
You mentioned
wavelength= 10*2= 20 cm=0.2m
velocity= 20m/s
frequency= velocity/wavelength=20/0.2=100 Hz
So i think they are given.. you wrote wavelength= 10*2= 20 cm=0.2m. Then why is it necessary to calculate wavelength again..You recalculated as wavelength=0.5*2= 1 m.

 

[Pranav Jha]

i calculate the wavelength as 0.2 m when the number of loops in the thread were 5 (to find out the frequency) and then recalculate it as 1 m (5*0.2) when there is only one loop ( to find out the fundamental harmonic frequency)

 

[PhysicsEnigma]

1. As the sound wave propagates away from the source, the energy is "thermally eaten". So, there is a gradual decrease in kinetic energy of the vibrating particles. Doesn't that lower the frequency of the sound wave? (i first considered from the classical energy point of view and then wondered if E=hv could be applied here? Please answer using both notions)

the frequency doesnot change. it always remains the same because the producer is not disturbed. it is only the ray or light that is decayed or changing its wavelength

 

[compitentboy]

Thanks for your quick replies. I was having a confusion with the formula f(n)=nf1

 

[SoniaAlissa]

Dear Rajini,

I could not get it that why wavelength to be calculated by multiplying by 2? Can you please help ?

Thanks
Alissa

 

[rbj]

E = h \nu is only for photons or light.

no, it's for more than photons and EM radiation.

 

[SoniaAlissa]

DEAR Mentor::The following is the Question and the solution provided by the members on the forum. Kindly help me understand that why we have calculated wavelength multiplying by 2? Thank you

Regards,
Alissa
------------------------------------------------------------------
A Stationary wave is making 5 loops. The distance between two consecutive statiopnary points is 10cm and its velocity is 20m/s.What will be its frequency?What will be the fundamental harmonic frequency?

SOLUTION:
wavelength= 10*2= 20 cm=0.2m
velocity= 20m/s
frequency= velocity/wavelength=20/0.2=100 Hz

I think:
for fundamental harmonic frequency, there should be only one loop along the whole string
Length of the string= 5*10=50cm=0.5m
so: wavelength=0.5*2= 1 m
frequency= 20/1= 20 Hz

 

[SoniaAlissa]

Dear rbj,

Thank you for the quick response. Actually I have got a little problem. The book I am following to help my student does not quote any such formula. I know the formula is correct, and everyone's use that, but can this be solved with some other method?

Waiting...
Thank you

Regards,
Alissa

 

[rbj]

Actually I have got a little problem. The book I am following to help my student does not quote any such formula. I know the formula is correct, and everyone's use that, but can this be solved with some other method?

it will come from a modern physics textbook. it's not in classical physics which says a wave is a wave and a particle is a particle.

but it turns out that sometimes a particle with energy E is also a wave of frequency \nu and vise versa. h is the conversion factor between the two.

i don't "solve" the formula, it just tell me how much energy particles of light (or whatever wave) have, given the frequency of the corresponding wave.

 

[SoniaAlissa]

it will come from a modern physics textbook. it's not in classical physics which says a wave is a wave and a particle is a particle.

but it turns out that sometimes a particle with energy E is also a wave of frequency \nu and vise versa. h is the conversion factor between the two.

i don't "solve" the formula, it just tell me how much energy particles of light (or whatever wave) have, given the frequency of the corresponding wave.

Please help me understand the conversion factor thing and as in the question it says that "the distance between TWO stationery points" is that a same thing? and is that represented by h which is valued as two.

Im sorry for bothering but I am trying to build the logic between two. Thank you

Regards,
Alissa

 

[Darwin123]

1. As the sound wave propagates away from the source, the energy is "thermally eaten". So, there is a gradual decrease in kinetic energy of the vibrating particles. Doesn't that lower the frequency of the sound wave? (i first considered from the classical energy point of view and then wondered if E=hv could be applied here? Please answer using both notions)

In quantum mechanics, the quasiparticle associated with a sound wave is called a phonon. The formula, E_Phonon=hv, is valid for the phonon where v is the frequency of the wave, h is Plancks constant, and E is the energy of a single phonon. However, a typical sound wave is made of trillions of phonons.
When a sound wave dissipates in classical physics, the amplitude of the wave decreases. When a sound wave dissipates in quantum mechanics, the phonon density corresponding to the wave decreases. The energy of each phonon doesn't usually change.
Let "n" be the number density of phonons corresponding to a wave of frequency v. The total energy density corresponding to the wave is:
E_Total=nhv,
where E_Total is the total energy density. Note that the total energy is the sum of the energies of the individual phonons.
There are two common misunderstanding regarding classical physics and quantum mechanics.
1) Some laymen think that classical mechanics doesn't allow discrete spacing of frequency.
-Classical physics allows the frequency of a standing wave to be "discretized".
-For instance, the modes of a plucked string with fixed endpoints has a quantized frequency even in classical physics.
2) Some laymen think that quantum mechanics is about quantization of frequency.
-In fact, quantum mechanics involves quantization of amplitude.

When a wave of a particular frequency loses or gains energy, it changes its amplitude. The frequency and wavelength do not have to change. The energy density of the wave is proportional to the square of the amplitude. The total energy of the wave has to change in discrete steps. However, this isn't because the energy of each particle changes in discrete steps. It means that the amplitude of the wave changes in discrete steps.
A change in amplitude for the wave corresponds to a change in number of the particles. The quantization of amplitude is inconsistent with classical physics. That is why Planck's theory raised eyebrows.
Planck hypothesized that the amplitude of the harmonic oscillator changed in steps. He did not imagine the quanta of energy as corresponding to changes in frequency of the harmonic oscillator. He assumed that the amplitude of each harmonic oscillator changed in steps that were proportional to the fundamental frequency of the harmonic oscillator.