 Susan Nall

2022-01-11

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? Esther Phillips

Step 1
Knowns
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We know that the following equation gives the average power carried by a sinusoidal wave on a string:
${P}_{avg}=\frac{1}{2}\sqrt{\mu F}{\omega }^{2}{A}^{2}$
(1)
And the following equation describes the relationship between a wave's frequency and angular speed:
$\omega =2\pi f$ (R.1)
Step 2
Given
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The mass of the wire is
$m=3.00g=0.003kg$
, its length is
l=80.0 cm=0.800 m
and the tension in it is $F=25.0N$ ;
The wave traveling along the wire has frequency of $f=120.0Hz$ and its amplitude is
$1.6mm=1.60×{10}^{-3}m$
Step 3
Calculations
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(A) We are aware that the mass per unit length corresponds to the linear mass density; Consequently, the wire's linear mass density is given as follows:
$\mu =\frac{m}{l}=\frac{0.003kg}{0.800m}=3.75×{10}^{-3}kg/m$
We substitute for f into relation (R.1), so we get:
$\omega =2\pi \left(120{s}^{-1}\right)=754{s}^{-1}$
Now, we plug our values for $\mu$, F, $\omega$ and A into equation (1), so we get:
${P}_{avg}=\frac{1}{2}\sqrt{\left(3.75×{10}^{-3}\right)×\left(25.0\right)}\cdot {\left(754\right)}^{2}\cdot \left(1.60×{10}^{-3}\right)$
$\therefore {P}_{avg}=0.223W$ Maria Lopez

Step 4
(b) We can observe from equation (1) that the average power is related to the amplitude's square;
${P}_{avg}\propto {A}^{2}⇒\frac{{P}_{avg,2}}{{P}_{avg,1}}={\left(\frac{{A}_{2}}{{A}_{1}}\right)}^{2}$
As a result, when the amplitude is cut in half, the average power is cut in half as well:
$\frac{{P}_{avg,2}}{{P}_{avg,1}}={\left(\frac{{A}_{\frac{1}{2}}}{{A}_{1}}\right)}^{2}=\frac{1}{4}$
${P}_{avg,2}=\frac{{P}_{avg}}{4}=\frac{0.223}{4}=0.056W$
$\therefore {P}_{avg,2}=0.056W$

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