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The power transmitted by a sinusoidal wave on a stretched string is $$P = \frac{1}{2}\mu v\omega^{2}A^{2}.$$ Is the power proportional to the square or the cube of frequency? $$v=\lambda f \:\:\: , \:\:\:\: \omega^{2}=(2\pi f)^{2}$$ $$ \Rightarrow \:\: P = 2\pi^{2}\mu\lambda A^{2}f^{3} $$ Why can't I consider the power proportional to the cube of frequency?

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In a resistor: Power =$I^{2}R$, but also Power =$\frac{V^2}{R}$. So is power proportional to $R$ or to $R^{-1}$ ? It depends what you keep constant: $I$ or $V$!

It's much the same in the case you're considering. If we're taking $v$ as a constant (which is pretty accurate for a given string under constant tension, as long as the wavelength isn't very short or the amplitude very large), then power is proportional to $f^2$.

The second version you consider would, with $\lambda$ held constant, give power proportional to $f^3$. But holding $\lambda$ constant while varying $f$ would be really weird as it would involve varying the string tension (in order to have different values of $v$ while leaving $\mu$ unchanged)!

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