91影视

The average power carried by a sinusoidal wave on a string is given by:

$P=\frac{1}{2}渭A^2{蝇}^{2}v$ --(1)

The speed of a periodic wave u with wavelength and frequency f is given by:

$v=f脳位$ --(2)

And the relation between the angular speed of a wave and its frequency is:

$w=2蟺f$--(3)

First, rearrange equation (2), so we get a relation between the frequency and the wavelength:

$f=\frac{v}{位}$

Substitute for f into relation (3), we get:

$蝇=\frac{2蟺v}{位}$

Now,substitute for w into equation (1), so we get:

$P_{avg}=\frac{1}{2}\sqrt{渭T}{\left(\frac{2蟺v}{位}\right)}^2{A}^2$

The average power carried by the wave is inversely proportional to the square of its wavelength:

role="math" localid="1664345158373" $$\begin{gathered} P_{avg}伪\frac{1}{{位}^2} \\ \frac{P_{avg,2}}{P_{avg,1}}=\frac{{{位}_1}^2}{{{位}_2}^2} \end{gathered}$$

So, when the wavelength is doubled ($位$鈧�= 2$位$鈧�), the average power becomes:

$$\begin{gathered} \frac{P_{avg,2}}{P_{avg,1}}=\frac{{{位}_1}^2}{{\left(2{位}_1\right)}^2} \\ =\frac{1}{4} \\ {P}_{avg,2}=\frac{P_{avg,1}}{4} \end{gathered}$$

Now, put in the value for P_avg,1 = 0.400W, so we get:

$$\begin{gathered} P_{avg,2}=\frac{0.400W}{4} \\ =0.100W \\ {P}_{avg,2}=0.100W \end{gathered}$$

Therefore, the average power carried by the waves is $P_{avg,2}=0.100W$.