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

15.11 The transverse displacement of a string (clamped at its both ends) is given by

y(x, t) = 0.06 sin( $\frac{2 π}{3} x)$ cos (120 $π t)$

where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0 ×10^–2 kg.

Answer the following :

(a) Does the function represent a travelling wave or a stationary wave?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions. What are the wavelength, frequency, and speed of each wave ?

(c) Determine the tension in the string

4 Views | Posted 5 months ago

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

(a) The general equation representing a stationary wave is given by the displacement function y(x,t) = 2asin kx cos $ω t$

The equation is similar to the give equation y(x, t) = 0.06 sin ( $\frac{2 π}{3} x)$ cos (120 $π t)$

Hence, the given function represents a stationary wave.

A wave travelling along the positive x-direction is given as: $y_{1}$ = asin ( $ω$ t – kx)

The wave travelling along the negative x-direction is given as: $y_{2}$ = asin ( $ω$ t + kx)

(b) The superposition of these two waves yields:

y = $y_{1} + y_{2}$ = asin ( $ω$ t – kx) + asin ( $ω$ t + kx)

= a sin( $ω$ t)cos(kx) – a sin

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(a) The general equation representing a stationary wave is given by the displacement function y(x,t) = 2asin kx cos $ω t$

The equation is similar to the give equation y(x, t) = 0.06 sin ( $\frac{2 π}{3} x)$ cos (120 $π t)$

Hence, the given function represents a stationary wave.

A wave travelling along the positive x-direction is given as: $y_{1}$ = asin ( $ω$ t – kx)

The wave travelling along the negative x-direction is given as: $y_{2}$ = asin ( $ω$ t + kx)

(b) The superposition of these two waves yields:

y = $y_{1} + y_{2}$ = asin ( $ω$ t – kx) + asin ( $ω$ t + kx)

= a sin( $ω$ t)cos(kx) – a sin(kx)cos( $ω$ t) – asin( $ω$ t)cos(kx) – asin(kx)cos( $ω$ t)

= -2a sin (kx) cos ( $ω$ t)

= -2asin( $\frac{2 π}{λ}$ x) cos (2 $π ν$ t) …….(i)

The transverse displacement of the string is given as

y(x, t) = 0.06 sin ( $\frac{2 π}{3} x)$ cos (120 $π t) \dots \dots \dots$ (ii)

Comparing equations (i) and (ii), we have $\frac{2 π}{λ}$ = $\frac{2 π}{3}$

Therefore $λ$ = 3m, it is given that 120 $π$ = 2 $π ν$ so Frequency, $ν = 60 H z$

Wave speed, v = $ν λ$ = 60 $\times 3$ = 180 m/s

The velocity of a transverse wave travelling in a string is given by the relation:

v = $\sqrt{\frac{T}{μ}}$ ….(i)

We have v = 180 m/s, mass of the string, m = 3 $\times 10^{- 2}$ kg,

length of the string, l = 1.5 m

Mass per unit length of the string, $μ$ = m/l = $\frac{3}{1.5} \times 10^{- 2}$ = $2 \times 10^{- 2}$ kg/m

(c) Tension in the string, T = $v^{2}$ $\times μ = 180^{2}$ $\times 2 \times 10^{- 2}$ = 648 N

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(a) The general equation representing a stationary wave is given by the displacement function y(x,t) = 2asin kx cos <math><mi>ω</mi><mi>t</mi></math>The equation is similar to the give equation y(x, t) = 0.06 sin ( <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>x</mi><mo>)</mo></math> cos (120 <math><mi>π</mi><mi>t</mi><mo>)</mo></math>Hence, the given function represents a stationary wave.A wave travelling along the positive x-direction is given as: <math><msub><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub></math> = asin ( <math><mi>ω</mi></math> t – kx)The wave travelling along the negative x-direction is given as: <math><msub><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = asin ( <math><mi>ω</mi></math> t + kx) (b) The superposition of these two waves yields:y = <math><msub><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>1</mn></mrow></mrow></msub><mo>+</mo><mi></mi><msub><mrow><mrow><mi>y</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msub></math> = asin ( <math><mi>ω</mi></math> t – kx) + asin ( <math><mi>ω</mi></math> t + kx)= a sin( <math><mi>ω</mi></math> t)cos(kx) – a sin(kx)cos( <math><mi>ω</mi></math> t) – asin( <math><mi>ω</mi></math> t)cos(kx) – asin(kx)cos( <math><mi>ω</mi></math> t)= -2a sin (kx) cos ( <math><mi>ω</mi></math> t)= -2asin( <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mi>λ</mi></mrow></mrow></mfrac></math> x) cos (2 <math><mi>π</mi><mi>ν</mi></math> t) …….(i)The transverse displacement of the string is given asy(x, t) = 0.06 sin ( <math><mfrac><mrow><mrow><mn>2</mn><mi>π</mi></mrow></mrow><mrow><mrow><mn>3</mn></mrow></mrow></mfrac><mi>x</mi><mo>)</mo></math> cos (120 <math><mi>π</mi><mi>t</mi><mo>)</mo><mo>…</mo><mo>…</mo><mo>…</mo></math> (ii)Comparing equations (i) and (ii), we have <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>3</mn></mrow></mrow></mfrac></math>Therefore <math><mi>λ</mi></math> = 3m, it is given that 120 <math><mi>π</mi></math> = 2 <math><mi>π</mi><mi>ν</mi></math> so Frequency, <math><mi>ν</mi><mo>=</mo><mn>60</mn><mi></mi><mi mathvariant="normal">H</mi><mi mathvariant="normal">z</mi></math>Wave speed, v = <math><mi>ν</mi><mi></mi><mi>λ</mi></math> = 60 <math><mo>×</mo><mn>3</mn></math> = 180 m/sThe velocity of a transverse wave travelling in a string 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> ….(i)We have v = 180 m/s, mass of the string, m = 3 <math><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> kg,length of the string, l = 1.5 mMass per unit length of the string, <math><mi>μ</mi></math> = m/l = <math><mfrac><mrow><mrow><mn>3</mn></mrow></mrow><mrow><mrow><mn>1.5</mn></mrow></mrow></mfrac><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> = <math><mn>2</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> kg/m (c) Tension in the string, T = <math><msup><mrow><mrow><mi>v</mi></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> <math><mo>×</mo><mi>μ</mi><mo>=</mo><mi></mi><msup><mrow><mrow><mn>180</mn></mrow></mrow><mrow><mrow><mn>2</mn></mrow></mrow></msup></math> <math><mo>×</mo><mn>2</mn><mo>×</mo><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>2</mn></mrow></mrow></msup></math> = 648 N

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