Light & Sound Waves: Thermal Energy & Frequency

In summary: So the restoring force will keep the frequency at a particular value, regardless of the amplitude. Am I understanding this correctly?In summary, amplitude and frequency are independent of each other.
  • #1
Pranav Jha
141
1
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)
 
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  • #2
E=hv is only for photons or light.
 
  • #3
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.
 
  • #4
The frequency must stay the same but the amplitude decays with distance.
 
  • #5
sophiecentaur said:
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?
 
  • #6
Pranav Jha said:
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.
 
  • #7
Pranav Jha said:
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
 
  • #8
Pranav Jha said:
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.
 
  • #9
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
 
  • #10
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.
 
  • #11
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
 
  • #12
Pranav Jha said:
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.
 
  • #13
Pranav Jha said:
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).
 
  • #14
Pranav Jha said:
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)..
 
  • #15
Rajini said:
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
 
  • #16
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.
 
  • #17
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)
 
  • #18
Pranav Jha said:
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 wavelenght
 
  • #19
Thanks for your quick replies. I was having a confusion with the formula f(n)=nf1
 
  • #20
Dear Rajini,

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

Thanks
Alissa
 
  • #21
Rajini said:
[itex] E = h \nu [/itex] is only for photons or light.

no, it's for more than photons and EM radiation.
 
  • #22
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
 
  • #23
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
 
  • #24
SoniaAlissa said:
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 [itex]E[/itex] is also a wave of frequency [itex]\nu[/itex] and vise versa. [itex]h[/itex] 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.
 
  • #25
rbj said:
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 [itex]E[/itex] is also a wave of frequency [itex]\nu[/itex] and vise versa. [itex]h[/itex] 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
 
  • #26
Pranav Jha said:
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.
 

1. What is the difference between light and sound waves?

Light waves are a type of electromagnetic radiation that can travel through a vacuum at a constant speed of 299,792,458 meters per second. Sound waves, on the other hand, are mechanical waves that require a medium, such as air or water, to travel through. They travel at a much slower speed of approximately 343 meters per second in air.

2. How is thermal energy related to light and sound waves?

Thermal energy is the energy associated with the motion of particles in a substance. Both light and sound waves can carry thermal energy, as they are made up of particles (photons for light waves and molecules for sound waves) that vibrate or move as the waves propagate. When these particles interact with other particles in a substance, they can transfer their thermal energy, causing a change in temperature.

3. What is the relationship between frequency and wavelength in light and sound waves?

Frequency and wavelength are inversely proportional in both light and sound waves. This means that as the frequency increases, the wavelength decreases, and vice versa. In light waves, frequency is directly related to the color of the light, with higher frequencies corresponding to shorter wavelengths and colors like blue and violet. In sound waves, frequency is directly related to the pitch of the sound, with higher frequencies corresponding to higher pitches.

4. How do light and sound waves interact with matter?

Light waves can interact with matter in three main ways: absorption, reflection, and transmission. When light waves are absorbed by matter, they transfer their energy to the particles in the substance, causing them to vibrate and increase in thermal energy. When light waves are reflected, they bounce off the surface of an object, allowing us to see the object. Sound waves also interact with matter through absorption and reflection, but they can also be transmitted through certain materials, such as air or water.

5. How are light and sound waves used in everyday life?

Light waves have a wide range of applications in everyday life, such as in lighting, communication (including radio, television, and the internet), and medical imaging. Sound waves are also used in a variety of ways, including communication (such as in telephones and speakers), navigation (like sonar and echolocation), and music and entertainment. Both light and sound waves are essential for our daily activities and have many practical uses in technology and medicine.

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