Question

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.

Answer 1

Answer:

(a) $P_{avg}$ = 0.22 W

(b) $P_{avg,2}$ = 0.056 W

Explanation:

given information:

the mass of piano wire, m = 3.00 g = 0.003 kg

tension, F = 25 N

length, l = 80 cm = 0.8 m

frequency, f = 120 Hz

amplitude, A = 1.6 mm = 0.0016 m

(a) the average power carried by the wave, $P_{avg}$

$P_{avg}$ = $\frac{1}{2}$(√μF)ω²A²

where,

ω = 2πf = 2π120 = 754$s^{-1}$

μ = $\frac{m}{l}$

  = $\frac{0.003}{0.8}$

  = 0.00375 kg/m

thus,

$P_{avg}$ = $\frac{1}{2}$(√(0.00375)(25))(754)²(0.0016)²

$P_{avg}$ = 0.22 W

(b) What happens to the average power if the wave amplitude is halved.

based on the equation above, we know that the average power is proportional to the square amplitude. therefore

$\frac{P_{avg,1} }{P_{avg,2} } = \frac{A_{1} ^{2} }{A_{2} ^{2} } }$

$\frac{P_{avg,1} }{P_{avg,2} } = \frac{A_{1} ^{2} }{(0.5A_{1} )^{2} } }$

$P_{avg,2}$ = $\frac{0.22}{4}$

$P_{avg,2}$ = 0.056 W

Answer 2

The average power carried by the wave is 0.2228 W and  if the wave amplitude is halved the average power will be 0.0557 W.

Given to us:

mass of piano wire, m = 3.00 g = 0.003 kg

length, l = 80.0 cm = 0.8 m

tension, F = 25.0 N

 frequency, f = 120.0 Hz

Amplitude, A = 1.6 mm = 0.0016 m

$\begin{aligned}\omega&= 2\pi f\\&=2\times\pi\times 120\\&=240\pi\ s^{-1}\end{aligned}$

$\begin{aligned}\mu&=\dfrac{m}{l} \\&=\frac{0.003}{0.8} \\&=0.00375\ kg/m\end{aligned}$

(a) To calculate the average power carried by the wave,

$P_{avg. = E \lambda T = \frac{1}{2}\mu \ A^2 \omega^2 \lambda\ T =\frac{1}{2} \mu\ A^2\omega^ 2 v= \frac{1}{2}(\sqrt{\mu f}) \ A^2 \omega^2$

Putting in all the values,

$\begin{aligned}P_{avg.} &= E \lambda T = \frac{1}{2}\mu \ A^2 \omega^2 \lambda\ T =\frac{1}{2} \mu\ A^2\omega^ 2 v= \frac{1}{2}(\sqrt{\mu F}) \ A^2 \omega^2\\P_{avg.} &= \frac{1}{2}(\sqrt{\mu F}) \ A^2 \omega^2\\&=\frac{1}{2} \times(\sqrt{0.00375\times25})\times0.0016^2\times(240\pi)^2\\&=0.2228\ W\end{aligned}$

Therefore, the average power carried by the wave is 0.2228 W.

b.) To calculate the average power if the wave amplitude is halved,

As we know that the average power is always proportional to the square of amplitude. therefore,

$\dfrac{P_{avg1}}{P_{avg2}} = (\dfrac{A_1}{A_2})^2 \\\\\\\dfrac{0.2228}{P_{avg2}} = (\dfrac{0.0016}{0.0008})^2 \\\\\\\dfrac{0.2228}{P_{avg2}} = (\dfrac{2}{1})^2 \\\\P_{avg.2}=\dfrac{0.2228}{4}\\\\P_{avg.2}=0.0557\ W$

Therefore, if the wave amplitude is halved the average power will be 0.0557 W.

Hence, The average power carried by the wave is 0.2228 W and  if the wave amplitude is halved the average power will be 0.0557 W.

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