15.10 For the travelling harmonic wave

**y(x, t) = 2.0 cos 2 (10t – 0.0080 x + 0.35)

where x and y are in cm and t in s. Calculate the phase difference between oscillatory motion of two points separated by a distance of

(a) 4 m

(b) 0.5 m

(c) $λ$ /2

(d)** 3 /4

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Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

The equation for a travelling harmonic wave is given by

y(x, t) = 2.0 cos 2 $π$ (10t – 0.0080 x + 0.35) or

= 2.0 cos (20 $π$ t – 0.016 $π$ x + 0.70 $π$ )

Comparing with classical equation y (x, t) = a $\sin ? (ω t + k x + φ)$ , we get

Amplitude a = 2.0 cm. Propagation constant, k = 0.016 $π$ , angular frequent, $ω = 20 π r a d / s$

From the relation $φ = k x = \frac{2 π}{λ} x$

(a) For x = 4 m = 400 cm, we have $φ$ = 0.016 $π \times 400$ = 6.4 $π$ rad

(b) For x = 0.5 m = 50 cm, we have $φ$ = 0.016 $π \times 50$ = 0.8 $π$ rad

(c) For x = $\frac{λ}{2}$ , we have $φ = \frac{2 π}{λ} \times$ $\frac{λ}{2}$ = $π$ rad

(d) For x = $\frac{3 λ}{4}$ , we have $φ = \frac{2 π}{λ} \times$ &n

...more

The equation for a travelling harmonic wave is given by

y(x, t) = 2.0 cos 2 $π$ (10t – 0.0080 x + 0.35) or

= 2.0 cos (20 $π$ t – 0.016 $π$ x + 0.70 $π$ )

Comparing with classical equation y (x, t) = a $\sin ? (ω t + k x + φ)$ , we get

Amplitude a = 2.0 cm. Propagation constant, k = 0.016 $π$ , angular frequent, $ω = 20 π r a d / s$

From the relation $φ = k x = \frac{2 π}{λ} x$

(a) For x = 4 m = 400 cm, we have $φ$ = 0.016 $π \times 400$ = 6.4 $π$ rad

(b) For x = 0.5 m = 50 cm, we have $φ$ = 0.016 $π \times 50$ = 0.8 $π$ rad

(c) For x = $\frac{λ}{2}$ , we have $φ = \frac{2 π}{λ} \times$ $\frac{λ}{2}$ = $π$ rad

(d) For x = $\frac{3 λ}{4}$ , we have $φ = \frac{2 π}{λ} \times$ $\frac{3 λ}{4}$ = $1.5 π$ rad

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The equation for a travelling harmonic wave is given byy(x, t) = 2.0 cos 2 <math><mi>π</mi></math> (10t – 0.0080 x + 0.35) or= 2.0 cos (20 <math><mi>π</mi></math> t – 0.016 <math><mi>π</mi></math> x + 0.70 <math><mi>π</mi></math> )Comparing with classical equation y (x, t) = a <math><mrow><mrow><mi mathvariant="normal">sin</mi></mrow><mo>?</mo><mrow><mo>(</mo><mi>ω</mi></mrow></mrow><mi>t</mi><mo>+</mo><mi>k</mi><mi>x</mi><mo>+</mo><mi></mi><mi>φ</mi><mo>)</mo></math> , we getAmplitude a = 2.0 cm. Propagation constant, k = 0.016 <math><mi>π</mi></math> , angular frequent, <math><mi>ω</mi><mo>=</mo><mn>20</mn><mi>π</mi><mi></mi><mi>r</mi><mi>a</mi><mi>d</mi><mo>/</mo><mi>s</mi></math>From the relation <math><mi>φ</mi><mo>=</mo><mi>k</mi><mi>x</mi><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac><mi>x</mi></math>(a) For x = 4 m = 400 cm, we have <math><mi>φ</mi></math> = 0.016 <math><mi>π</mi><mi></mi><mo>×</mo><mn>400</mn><mi></mi></math> = 6.4 <math><mi>π</mi></math> rad (b) For x = 0.5 m = 50 cm, we have <math><mi>φ</mi></math> = 0.016 <math><mi>π</mi><mi></mi><mo>×</mo><mn>50</mn><mi></mi></math> = 0.8 <math><mi>π</mi></math> rad (c) For x = <math><mfrac><mrow><mrow><mi>λ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math> , we have <math><mi>φ</mi><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac><mo>×</mo></math> <math><mfrac><mrow><mrow><mi>λ</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></mfrac></math> = <math><mi>π</mi></math> rad (d) For x = <math><mfrac><mrow><mrow><mn>3</mn><mi>λ</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac></math> , we have <math><mi>φ</mi><mo>=</mo><mi></mi><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac><mo>×</mo></math> <math><mfrac><mrow><mrow><mn>3</mn><mi>λ</mi></mrow></mrow><mrow><mrow><mn>4</mn></mrow></mrow></mfrac></math> = <math><mn>1.5</mn><mi>π</mi></math> rad