Question

If mechanical waves A and B are traveling in the same medium and transfer the same amount of energy per unit time, but wave A has an amplitude z times the amplitude of wave B, what is the proportion of their wavelengths?

Answer 1

I would think that the answer to this would be \[\frac{ \lambda_{A} }{ \lambda_{B} } = 1/z^2\], since the energy carried by a wave is proportional to the square of its amplitude but this is apparently incorrect.

Answer 2

Energy carried by a vibrating string per unit of time, namely the intensity is given by the subsequent formula:
\[I=\frac{ 1 }{ 2 }\mu *v *\omega ^{2}*A ^{2}\]
where mu is the linear density of mass, v is the speed of the wave, omega is the angular speed of the wave and A is the amplitude of the wave.
So, from the text of your problem we can write:
\[v _{1}\omega _{1}^{2}A _{1}^{2}=v _{2}\omega _{2}^{2}A _{2}^{2}\]
and:
\[\lambda _{1}f _{1}^{3}z ^{2}=\lambda _{2}f _{2}^{3}\]
being \[v=\lambda*f,\quad \omega=2*\pi*f\]
f is the frequency of the wave

Answer 3

Yes. However, aren't the wave speeds of A and B equal as they are traveling on the same medium?

Answer 4

yes they are

Answer 5

so we have:
\[\lambda _{1}f _{1}=\lambda _{2}f _{2}\]
and finally:
\[z ^{2}=\left( \frac{ \lambda _{1} }{ \lambda _{2} } \right)^{2}\]

Answer 6

is it your result?

Answer 7

I do not follow your last step. I understand the first relation, but I don't see z^2 = (lambda1/lambda2)^2

Answer 8

Please note that I can write this:
\[\lambda _{1}f _{1}^{3}z ^{2}=(\lambda _{1}f _{1})*f _{1}^{2}*z ^{2}\]
and:
\[\lambda _{2}f _{2}^{3}=(\lambda _{2}f _{2})*f _{2}^{2}\]
so:
\[f _{1}^{2}*z ^{2}=f _{2}^{2}\]
furthermore, we have:
\[\frac{ \lambda _{1} }{ \lambda _{2} }=\frac{ f _{2} }{ f _{1} }\]

Answer 9

Ahhh. I see now. thank you