A piano wire with mass 3.00 g and length 80.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplitude 1.6 mm travels along the wire. (a) Calculate the average power carried by the wave. (b) What happens to the average power if the wave amplitude is halved?

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

$P_{avg}=\frac{1}{2}\sqrt{\mu T}{\omega }^2{A}^2$ --(1)

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

$\omega =2\pi f$ --(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:
$$\begin{gathered} \mu =\frac{m}{l} \\ =\frac{0.003kg}{0.800m} \\ =3.75\times {10}^{-3}kg/m \end{gathered}$$

We substitute for f into relation (2):
$$\begin{gathered} \omega =2\pi \left(120s^{-1}\right) \\ =754s^{-1} \end{gathered}$$

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

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

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

$P_{avg}\alpha A^2$

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