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Expert-verified Found in: Page 474 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # A sinusoidal wave of angular frequency 1200 rad/s and amplitude 3,00 mm is sent along a cord with linear density 2.00 g/m and tension 1200 N. (a)What is the average rate at which energy is transported by the wave to the opposite end of the cord? (b)If, simultaneously, an identical wave travels along an adjacent, identical cord, what is the total average rate at which energy is transported to the opposite ends of the two cords by the waves?If, instead, those two waves are sent along the same cord simultaneously, what is the total average rate at which they transport energy When their phase difference is 0, (b)When their phase difference is (c) 0(d) ${\mathbf{0}}{\mathbf{.}}{\mathbf{4}}{\mathbf{\pi }}{\mathbit{r}}{\mathbit{a}}{\mathbit{d}}$, and (e) is ${\mathbf{\pi }}{\mathbit{r}}{\mathbit{a}}{\mathbit{d}}$?

1. The average rate at which energy is transported by the wave to the opposite end of the cord is 10 W
2. The total average rate at which energy is transported to the opposite ends of the two cords by the waves is 20 W
3. The total average rate at which they transport energy when their phase difference is zero is 40 W
4. The total average rate at which they transport energy when their phase difference is is 26 W
5. The total average rate at which they transport energy when their phase difference is is 0 W
See the step by step solution

## Step 1: The given data

1. Angular frequency of wave, $\omega =1200\mathrm{rad}/\mathrm{s}$
2. Amplitude of wave, ${y}_{m}=3\mathrm{mm}\mathrm{or}3×{10}^{-3}\mathrm{m}$
3. Liner density of wave, $\mu =2.00\mathrm{g}/\mathrm{m}\mathrm{or}2.00×{10}^{-3}\mathrm{kg}/\mathrm{m}$
4. Tension in string, T = 1200 N

## Step 2: Understanding the concept of energy

When we set up a wave on a stretched string, we provide energy for the motion of the string. Therefore, energy is transported in the form of both kinetic as well as potential energy. From the speed of the wave, we can calculate the average power of the wave in various situations, that is, the average rate at which energy is transported by the wave.

Formula:

The velocity of a string, $v=\sqrt{T/\mu }$ (i)

Average power or energy of a body, ${P}_{ave}=\frac{1}{2}\mu v{\omega }^{2}{y}_{m}^{2}$ (ii)

Amplitude of a wave, $A=2{y}_{m}\mathrm{cos}\frac{\mathrm{\varphi }}{2}$ (iii)

## Step 3: a) Calculation of average rate at which energy is transported to the opposite end

Using equation (i) and the given values, we get the velocity of the wave as:

$v=\sqrt{\frac{1200}{2×{10}^{-3}}}\phantom{\rule{0ex}{0ex}}\sqrt{600×{10}^{3}}\phantom{\rule{0ex}{0ex}}=775\mathrm{m}/\mathrm{s}$

Using this speed in the equation (ii), we can get the average energy as:

role="math" localid="1661157835362" ${P}_{ave}=\frac{1}{2}×2×{10}^{-3}×775×{1200}^{2}×{\left(3×{10}^{-3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=775×144×{10}^{4}×{10}^{-3}×9×{10}^{-6}\phantom{\rule{0ex}{0ex}}=10\mathrm{W}$

Hence, the value of average energy rate is 10 W.

## Step 4: b) Calculation of total average energy rate at which energy is transported to the two opposite cords by the waves

If an identical wave travels along an adjacent, identical cord, then the average rate at which energy is transported is twice of that single wave, that is,. No superposition occurs here.

Speed of wave is v = 775 m/s

Hence, using equation (ii), the total average energy rate is given as:

${P}_{ave}=2×\frac{1}{2}×2×{10}^{-3}×775×{1200}^{2}×{\left(3×{10}^{-3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=2×775×144×{10}^{4}×{10}^{-3}×9×{10}^{-6}\phantom{\rule{0ex}{0ex}}=20\mathrm{W}$

Hence, the value of total average rate by the waves is 20 W

## Step 5: c) Calculation of the total average rate at which they transport energy when their phase difference is zero

If the phase difference between the two waves is i.e.the two waves will haveconstructive interference,the new amplitude of the wave using equation (iii) is given as:

$A=2{y}_{m}\mathrm{cos}\frac{0}{2}\phantom{\rule{0ex}{0ex}}=2{y}_{m}\phantom{\rule{0ex}{0ex}}=6×{10}^{-3}\mathrm{m}$

Then, using equation (i), the average rate at which energy is transported is given as:

${P}_{ave}=\frac{1}{2}×2×{10}^{-3}×775×{1200}^{2}×{\left(6×{10}^{-3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=775×144×{10}^{4}×{10}^{-3}×36×{10}^{-6}\phantom{\rule{0ex}{0ex}}=40\mathrm{W}$

Hence, the total average rate when phase is zero is 40 W

## Step 6: d) Calculation of the total average rate at which they transport energy when their phase difference is 4πrad

Phase difference between two waves is $0.4\mathrm{\pi }$ then the new amplitude of the wave using equation (iii) is given as:

$A=2{y}_{m}\mathrm{cos}\frac{0.4\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}=2{y}_{m}\mathrm{cos}0.2\mathrm{\pi }\phantom{\rule{0ex}{0ex}}=2×{\mathrm{y}}_{\mathrm{m}}×0.8090\phantom{\rule{0ex}{0ex}}=1.618{\mathrm{y}}_{\mathrm{m}}$

From that we can find new average rate of energy transport using equation (i) is given as:

role="math" localid="1661162817509" ${P}_{ave}=\frac{1}{2}\mu v{\omega }^{2}{\left(1.618{y}_{m}\right)}^{2}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}×2×{10}^{-3}×775×{1200}^{2}×{\left(1.618\right)}^{2}×{\left(3×{10}^{-3}\right)}^{2}\phantom{\rule{0ex}{0ex}}=775×144×{10}^{4}×{10}^{-3}×2.618×9×{10}^{-6}\phantom{\rule{0ex}{0ex}}=26\mathrm{W}$

Hence, the value of the total average energy rate is 26W

## Step 7: e) Calculation of the total average rate at which they transport energy when their phase difference is πrad

Phase difference between two waves is then the new amplitude of the wave using equation (iii), we get

$A=2{y}_{m}\mathrm{cos}\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}=2{y}_{m}×0\phantom{\rule{0ex}{0ex}}=0\phantom{\rule{0ex}{0ex}}$

From that we can find new average rate of energy transport using equation (i), we get

${P}_{ave}=\frac{1}{2}\mu \mu v{\omega }^{2}{A}^{2}\phantom{\rule{0ex}{0ex}}=0\mathrm{W}$

Hence, the value of the total average rate of energy is 0 W ### Want to see more solutions like these? 