# Light and sound waves

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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E=hv is only for photons or light.

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

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
Gold Member
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
Gold Member
2019 Award
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 doesnt 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 dont 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
Gold Member
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.

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

Can any one 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.

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
Mentor
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
Gold Member
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).

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)..

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

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.

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)

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

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

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.

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

Dear rbj,

Thank you for the quick response. Actually I have got a little problem. The book Im 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 Im 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.

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 Im trying to build the logic between two. Thank you

Regards,
Alissa