91Ó°ÊÓ

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

$P_{avg}=\frac{1}{2}\sqrt{\mathrm{μ}T}{\mathrm{Ó\neg }}^2{A}^2$ --(1)

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

$\mathrm{Ó\neg }=2\mathrm{Ï€}f$ --(2)

The mass of the wire is m = 3.00 g = 0.003 kg, its length is l = 80.0 cm = 0.800 m and the tension in it is T = 25.0 N;

The wave traveling along the wire has frequency of f 120.0 Hz and its amplitude is 1.6 mm = 1.60 x 10-³ m.

(a) The linear mass density is the mass per unit length. So the linear mass density of the wire is:
$$\begin{gathered} \mathrm{μ}=\frac{m}{l} \\ =\frac{0.003kg}{0.800m} \\ =3.75×{10}^{-3}kg/m \end{gathered}$$

We substitute for f into relation (2):
$$\begin{gathered} \mathrm{Ó\neg }=2\mathrm{Ï€}\left(120s^{-1}\right) \\ =754s^{-1} \end{gathered}$$

Now, putting in the values for, F, w and A into equation (1), so we get:

$$\begin{gathered} P_{avg}=\frac{1}{2}\sqrt{\left(3.75×{10}^{-3}\right)×\left(25.0\right)}{\left(754\right)}^2\left(1.60×{10}^{-3}\right) \\ {P}_{avg}=0.223W \end{gathered}$$

(b) From equation (1), the average power is proportional to the square of the amplitude;

$P_{avg}\mathrm{Î\pm }{A}^2$

So, when the amplitude is reduced to half of its value, the average power is reduced to quarter of its value:

$$\begin{gathered} \frac{P_{avg,2}}{P_{avg,1}}={\left(\frac{A_1/2}{A_1}\right)}^2 \\ =\frac{1}{4} \\ {P}_{avg,2}=\frac{P_{avg}}{4} \\ =\frac{0.223}{4} \\ =0.056W \\ {P}_{avg,2}=0.056W \end{gathered}$$

Therefore, the average power carried by the wave is P_avg = 0.223 W and the average power of the wave if the wave amplitude is halved is P_avg,2 = 0.056 W