Question

A light wire is tightly stretched with tension F. Transverse traveling waves of amplitude A and wavelength A₁carry average power P_avg = 0.400 W. If the wavelength of the waves is doubled, so A₂= 2A₁, while the tension F and amplitude A are not altered, what then is the average power Pav,2 carried by the waves?

Short answer

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

Step-by-step solution

Determination of the formula of Mechanical Waves

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

$P=\frac{1}{2}\mu A^2{\omega }^{2}v$ --(1)

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

$v=f\times \lambda$ --(2)

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

$w=2\pi f$--(3)

Step 2:Application of the formula of Mechanical Waves

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

$f=\frac{v}{\lambda }$

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

$\omega =\frac{2\pi v}{\lambda }$

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

$P_{avg}=\frac{1}{2}\sqrt{\mu T}{\left(\frac{2\pi v}{\lambda }\right)}^2{A}^2$

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

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

So, when the wavelength is doubled ($\lambda$₂= 2$\lambda$₁), the average power becomes:

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

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 $P_{avg,2}=0.100W$.