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E=hv is only for photons or light.

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sophiecentaur

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

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Why is the frequency of a wave unaffected by its amplitude? Isn't it kind of counter-intuitive?The frequency must stay the same but the amplitude decays with distance.

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Born2bwire

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

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davenn

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

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

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sophiecentaur

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

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But it obviously is wrong! Please clear up my thinking in this regard

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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.

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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

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russ_watters

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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.i am not familiar with simple harmonic motion.

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sophiecentaur

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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.

But it obviously is wrong! Please clear up my thinking in this regard

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).

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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)..

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

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what do you mean? i don't think they are givenI 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)..

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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.

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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

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

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

Thanks

Alissa

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no, it's for more than photons and EM radiation.[itex] E = h \nu [/itex] is only for photons or light.

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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

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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

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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.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?

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.

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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.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.

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

Regards,

Alissa

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