What is the relationship between amplitude and frequency of a wave? Some say there isn't any relationship, some say that there is, but from their answers the relationship is still unclear.
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$\begingroup$ Have you read the definitions? : en.wikipedia.org/wiki/Wave $\endgroup$– DavidmhCommented May 18, 2014 at 19:31
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$\begingroup$ yes and the answer for my question in not there. $\endgroup$– MichaelCommented May 18, 2014 at 19:34
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3 Answers
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In general there is no relationship. Any combinations of frequencies and amplitudes are allowed.
There can be some relationship in certain special cases: for example if you have a source of waves which emits a specific spectrum, then the amplitudes and frequencies obey that spectrum. But spectrums can be arbitrary, so the dependency can be arbitrary.
In conclusion: generally there is no relationship.
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$\begingroup$ If a source of sound waves emits a specific spectrum, in that what would be the relation between amplitude and frequency? $\endgroup$– rainmanCommented Oct 10, 2014 at 9:31
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If the proposed wave can be represented as sinusoidal and moving in the $+x$ direction, then that means $y(x,t)=Acos(kx-\omega t)$, where $A$=Amplitude, $k$=wave number, $x$=horizontal direction, $\omega$=angular velocity, $t$=time, the derivation of which can be obtained from Young and Freedman's modern physics 14th edition. Now, the first partial derivative of the position function, $y(x,t)$ yields the velocity function $v(x,t)=-\omega Acos(kx-\omega t)$. A pivotal substitution for $ω=2\pi f$ yields $v(x,t)=-2\pi fAcos(kx-2πft)$. Although some attenuation of signal occurs between source as listener, wave velocity is generally constant, and therefore, when the $cos$ is at maximum of $1$, then velocity is also maximized. Therefore substituting $1$ for cos yields maximum velocity at $v(x,t)=2\pi fA$. Solving for Amplitude, we have $A=v(x,t)/2πf$, which directly allows the calculation of amplitude given frequency where $v(x,t)_{sound} = 344 m/s$ at $20^0C$ and $v(x,t)_{light}=3.00\times10^8 m/s.$
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$\begingroup$ @LeonardMartin-I think your $y(x,t)$ function describes the vertical displacement, and so the velocity you refer to is the velocity in the vertical direction, which after taking the derivative is a $sin$-function. This means the magnitude of the vertical velocity is maximum when the $sin$ is equal to one. This happens at values of $y(x,t)$ where the vertical displacement is zero. $\endgroup$ Commented Mar 7, 2017 at 15:12
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Planck's function ($E=hv$) applied to $E∝A^2$. Light has fixed qualities making its boundary conditions simple. Those can easily be altered by the medium of propagation or the nature of the wave (e.g. sound). So generally speaking, no relationship is assumed, but in specific applications, relationship can be found by establishing the working boundaries. That's why you get mixed feedback on this question.
