15.24 One end of a long string of linear mass density 8.0 × 10^–3 kg m^–1 is connected to an electrically driven tuning fork of frequency 256 Hz. The other end passes over a pulley and is tied to a pan containing a mass of 90 kg. The pulley end absorbs all the incoming energy so that reflected waves at this end have negligible amplitude. At t = 0, the left end (fork end) of the string x = 0 has zero transverse displacement (y = 0) and is moving along positive y-direction. The amplitude of the wave is 5.0 cm. Write down the transverse displacement y as function of x and t that describes the wave on the string.

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Vishal Baghel | Contributor-Level 10

5 months ago

The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $ω t - k x)$ ……… (i)

Linear mass density $μ =$ 8.0 $\times 10^{- 3}$ kg/m and frequency of the tuning fork, $ν$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $\times 9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ = $\sqrt{\frac{882}{8.0 \times 10^{- 3}}}$ = 332 m/s

Angular frequency, $ω$ = 2 $π ν$ = 2 $π \times 256$ = 1608.5 rad/s = 1.6 $\times 10^{3}$ rad/s

Wavelength, $λ$ =

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The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:

y (x, t) = a sin ( $ω t - k x)$ ……… (i)

Linear mass density $μ =$ 8.0 $\times 10^{- 3}$ kg/m and frequency of the tuning fork, $ν$ = 256 Hz

Amplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)

Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 $\times 9.8$ = 882 N

The velocity of the transverse wave, v is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ = $\sqrt{\frac{882}{8.0 \times 10^{- 3}}}$ = 332 m/s

Angular frequency, $ω$ = 2 $π ν$ = 2 $π \times 256$ = 1608.5 rad/s = 1.6 $\times 10^{3}$ rad/s

Wavelength, $λ$ = $\frac{v}{ν}$ = $\frac{332}{256}$ = 1.297 m

Propagation constant, k = $\frac{2 π}{λ}$ = $\frac{2 π}{1.297}$ = 4.844 /m

Substituting these values in the displacement equation, we get

y (x, t) = a sin ( $ω t - k x)$

y (x, t) = 0.05 sin (1.6 $\times 10^{3}$ t – 4.844 x) m

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The equation of a travelling wave propagating along the positive y- direction is given by the displacement equation:y (x, t) = a sin ( <math><mi>ω</mi><mi>t</mi><mo>-</mo><mi>k</mi><mi>x</mi><mo>)</mo></math> ……… (i)Linear mass density <math><mi>μ</mi><mo>=</mo><mi></mi></math> 8.0 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></math> kg/m and frequency of the tuning fork,  <math><mi>ν</mi></math> = 256 HzAmplitude of the wave, a= 5.0 cm = 0.05 m …. (ii)Mass of the pan, m = 90 kg and tension of the string, T = mg = 90 <math><mo>×</mo><mn>9.8</mn></math> = 882 NThe velocity of the transverse wave, v is given by the relation:v = <math><msqrt><mrow><mfrac><mrow><mrow><mi>T</mi></mrow></mrow><mrow><mrow><mi>μ</mi></mrow></mrow></mfrac></mrow></msqrt></math> = <math><msqrt><mrow><mfrac><mrow><mrow><mn>882</mn></mrow></mrow><mrow><mrow><mn>8.0</mn><mi mathvariant="normal"></mi><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>3</mn></mrow></mrow></msup></mrow></mrow></mfrac></mrow></msqrt></math> = 332 m/sAngular frequency,  <math><mi>ω</mi></math> = 2 <math><mi>π</mi><mi>ν</mi></math> = 2 <math><mi>π</mi><mi></mi><mo>×</mo><mn>256</mn></math> = 1608.5 rad/s = 1.6 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> rad/sWavelength,  <math><mi>λ</mi></math> = <math><mfrac><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mi>ν</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>332</mn></mrow></mrow><mrow><mrow><mn>256</mn></mrow></mrow></mfrac></math> = 1.297 mPropagation constant, k = <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mn>1.297</mn></mrow></mrow></mfrac></math> = 4.844 /mSubstituting these values in the displacement equation, we gety (x, t) = a sin ( <math><mi>ω</mi><mi>t</mi><mo>-</mo><mi>k</mi><mi>x</mi><mo>)</mo></math>y (x, t) = 0.05 sin (1.6 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></msup></math> t – 4.844 x) m

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