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Waves physics ncert solutions class 11th Physics Class 11th Physics Waves

15.22 A travelling harmonic wave on a string is described by

y(x, t) = 7.5 sin (0.0050x +12t + $π$ /4)

(a) What are the displacement and velocity of oscillation of a point at x = 1 cm, and t = 1 s? Is this velocity equal to the velocity of wave propagation?

(b) Locate the points of the string which have the same transverse displacements and velocity as the x = 1 cm point at t = 2 s, 5 s and 11 s.

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Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

(a) The given harmonic wave is

y(x, t) = 7.5 sin (0.0050x +12t + $π$ /4)

For x = 1 cm and t = 1 s

y(1, 1) = 7.5 sin (0.0050 +12+ $π$ /4)

= 7.5 sin( 12.0050 + $\frac{π}{4}$ )

= 7.5 sin (12.0050 + 0.7854)

= 7.5 sin (732.84 $°$ )

= 7.5 sin (90 $\times 8 +$ 12.81) $°$

= 7.5 sin 12.81 $°$

=1.6629 cm

The velocity of the oscillation at a given point and the time is given as:

v = $\frac{d}{d t}$ y(x,t) = $\frac{d}{d t} [7.5 s i n (0.0050 x + 12 t + π / 4)]$

= 7.5 $\times 12 c o s ? (0.0050 x + 12 t + π / 4$ )

At x= 1 cm and t= 1 s

v= y (1,1) = 90 cos (12.005 + $\frac{π}{4}$ ) = 90 cos(12.81 $°$ ) = 87.75 cm/s

Now the equation of a propagating wave is given by

Y(x, t) = a sin (kx + $ω$ t + $φ)$ , where

k =&nbs

...more

(a) The given harmonic wave is

y(x, t) = 7.5 sin (0.0050x +12t + $π$ /4)

For x = 1 cm and t = 1 s

y(1, 1) = 7.5 sin (0.0050 +12+ $π$ /4)

= 7.5 sin( 12.0050 + $\frac{π}{4}$ )

= 7.5 sin (12.0050 + 0.7854)

= 7.5 sin (732.84 $°$ )

= 7.5 sin (90 $\times 8 +$ 12.81) $°$

= 7.5 sin 12.81 $°$

=1.6629 cm

The velocity of the oscillation at a given point and the time is given as:

v = $\frac{d}{d t}$ y(x,t) = $\frac{d}{d t} [7.5 s i n (0.0050 x + 12 t + π / 4)]$

= 7.5 $\times 12 c o s ? (0.0050 x + 12 t + π / 4$ )

At x= 1 cm and t= 1 s

v= y (1,1) = 90 cos (12.005 + $\frac{π}{4}$ ) = 90 cos(12.81 $°$ ) = 87.75 cm/s

Now the equation of a propagating wave is given by

Y(x, t) = a sin (kx + $ω$ t + $φ)$ , where

k = $\frac{2 π}{λ}$ or $λ$ = $\frac{2 π}{k}$ and $ω = 2 π v$

Speed, v = $\frac{ω}{k}$ , where $ω$ = 12 rad/s and k = 0.0050 /m

v = $\frac{12}{0.0050}$ = 2400 cm/s

Hence, the velocity of the wave oscillation at x=1 cm and t=1s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as:

$λ$ = $\frac{2 π}{k}$ = $\frac{2 π}{0.0050}$ = 1256 cm = 12.56 m

Therefore, all the points at distances n $λ$ (n = $\pm 1, \pm 2 \dots \dots . a n d s o o n)$ i.e. $\pm 12.56 m, \pm 25.12 m \dots \dots .$ and so on for x=1 cm, will have the same displacement as the x=1 cm points at t= 2s, 5s and 11s.

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(a) The given harmonic wave isy(x, t) = 7.5 sin (0.0050x +12t + <math><mi>π</mi></math> /4)For x = 1 cm and t = 1 sy(1, 1) = 7.5 sin (0.0050 +12+ <math><mi mathvariant="normal">π</mi></math> /4)= 7.5 sin( 12.0050 + <math><mfrac><mrow><mrow><mi>π</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac></math> )= 7.5 sin (12.0050 + 0.7854)= 7.5 sin (732.84 <math><mo>°</mo></math> )= 7.5 sin (90 <math><mo>×</mo><mn>8</mn><mo>+</mo><mi></mi></math> 12.81) <math><mo>°</mo></math>= 7.5 sin 12.81 <math><mo>°</mo></math>=1.6629 cmThe velocity of the oscillation at a given point and the time is given as:v = <math><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac></math> y(x,t) = <math><mfrac><mrow><mrow><mi>d</mi></mrow></mrow><mrow><mrow><mi>d</mi><mi>t</mi></mrow></mrow></mfrac><mfenced open="[" close="]" separators="|"><mrow><mrow><mn>7.5</mn><mi mathvariant="normal"></mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">i</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal"></mi><mo>(</mo><mn>0.0050</mn><mi>x</mi><mi></mi><mo>+</mo><mn>12</mn><mi>t</mi><mi></mi><mo>+</mo><mi></mi><mi>π</mi><mo>/</mo><mn>4</mn><mo>)</mo></mrow></mrow></mfenced></math>= 7.5 <math><mo>×</mo><mn>12</mn><mi></mi><mi mathvariant="normal">c</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">s</mi><mo>?</mo><mo>(</mo><mn>0.0050</mn><mi>x</mi><mi></mi><mo>+</mo><mn>12</mn><mi>t</mi><mi></mi><mo>+</mo><mi></mi><mi>π</mi><mo>/</mo><mn>4</mn></math> )At x= 1 cm and t= 1 sv= y (1,1) = 90 cos (12.005 + <math><mfrac><mrow><mrow><mi>π</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac></math> ) = 90 cos(12.81 <math><mo>°</mo></math> ) = 87.75 cm/sNow the equation of a propagating wave is given byY(x, t) = a sin (kx + <math><mi>ω</mi></math> t + <math><mi>φ</mi><mo>)</mo></math> , wherek = <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac></math> or <math><mi>λ</mi></math> = <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>k</mi></mrow></mrow></mfrac></math> and <math><mi>ω</mi><mo>=</mo><mn>2</mn><mi>π</mi><mi>v</mi></math>Speed, v = <math><mfrac><mrow><mrow><mi>ω</mi></mrow></mrow><mrow><mrow><mi>k</mi></mrow></mrow></mfrac></math> , where <math><mi>ω</mi></math> = 12 rad/s and k = 0.0050 /mv = <math><mfrac><mrow><mrow><mn>12</mn></mrow></mrow><mrow><mrow><mn>0.0050</mn></mrow></mrow></mfrac></math> = 2400 cm/sHence, the velocity of the wave oscillation at x=1 cm and t=1s is not equal to the velocity of the wave propagation. (b) Propagation constant is related to wavelength as:<math><mi>λ</mi></math> = <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>k</mi></mrow></mrow></mfrac></math> = <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mn>0.0050</mn></mrow></mrow></mfrac></math> = 1256 cm = 12.56 mTherefore, all the points at distances n <math><mi>λ</mi></math> (n = <math><mo>±</mo><mn>1</mn><mo>,</mo><mi></mi><mo>±</mo><mn>2</mn><mo>…</mo><mo>…</mo><mo>.</mo><mi mathvariant="normal">a</mi><mi mathvariant="normal">n</mi><mi mathvariant="normal">d</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">s</mi><mi mathvariant="normal">o</mi><mi mathvariant="normal"></mi><mi mathvariant="normal">o</mi><mi mathvariant="normal">n</mi><mo>)</mo></math> i.e. <math><mo>±</mo><mi></mi><mn>12.56</mn><mi></mi><mi>m</mi><mo>,</mo><mi></mi><mo>±</mo><mn>25.12</mn><mi>m</mi><mo>…</mo><mo>…</mo><mo>.</mo></math> and so on for x=1 cm, will have the same displacement as the x=1 cm points at t= 2s, 5s and 11s.

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