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Q34P

Expert-verifiedFound in: Page 474

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**

- The average rate at which energy is transported by the wave to the opposite end of the cord is 10 W
- The total average rate at which energy is transported to the opposite ends of the two cords by the waves is 20 W
- The total average rate at which they transport energy when their phase difference is zero is 40 W
- The total average rate at which they transport energy when their phase difference is is 26 W
- The total average rate at which they transport energy when their phase difference is is 0 W

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

**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)

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

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

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

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

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

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:

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

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

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:

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

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

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

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

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

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

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

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

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

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

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

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

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