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To practice Problem-Solving Strategy 15.2 Standing Waves.In lab, your instructor generates a standing wave using a thin string of length L = 2.25 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 14.5 m/s . The first antinode of the standing wave is a distance of x = 37.5 cm from the left end of the string, while a light bead is placed a distance of 18.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.To check your equation for the standing wave's transverse velocity, find the maximum transverse velocity at x = 75.0 cm .​

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Answer:

Explanation:

Let the equation of standing wave be as follows

y = A sinωt cos kx

A = 2.45 mm

y = 2.45 cosωt sin kx

given

[tex]\frac{\omega}{k}[/tex]  = velocity = 14.5

Position of first antinode = 37.5 cm

kx = π / 2

k x 37.5 = π / 2

k = π / 75

ω / k = 14.5

ω = 14.5 x π / 75

= .607 rad /s

Maximum transverse speed

= ω A

=  .607 x 2.45

= 1.49 mm / s

y = A sinωt cos kx

Transverse velocity

v = dy / dt

= ω A cosωt cos kx

Maximum transverse velocity at any x = ω A .

The characteristics of standing waves allows find the results for the questions about the speed of the rope are:

  • The transverse oscillatory velocity is: vy = 0.298 m / s

Given parameters

  • The amplitude of the wave A = 2.45 mm = 2.45 10-3 m
  • Chord length L = 2.25 m
  • Wave velocity v = 14.5 m / d
  • The first antinode x = 37.5 cm

To find

  • Maximum rope swing speed.

The movement in a string is formed by two movements, a movement in the direction of the string with constant speed and a transverse movement where the speed varies as in a simple harmonic movement.

The standing wave is formed from the sum of the incident wave and the reflected wave.

         y = A cos (kx- wt)

         y = A cos (kx + wt)

resulting

       y = A sin wt cos kx

the speed that the wave is given by

       v = w / k

They indicate the position of the first antinode at this point the cosine function must be maximum.

     kx = π

     k = π/x

     k = [tex]\frac{\pi }{0.375}[/tex]  

     k = 8.38 m⁻¹

let's find the angular velocity.

     w = v k

     w = 14.5  8.38

     w = 121.5 rad / s

     

The expression for displacement in simple harmonic motion is:

     x = A cos wt

The speed is defined by the variation of the position with respect to time.

      v = [tex]\frac{dx}{dt}[/tex] =

       v = - A w sin wt

To calculate the maximum speed we make the sine equal to 1.

      [tex]v_{y \ max}[/tex]  = w A

      [tex]v_{y \ max}[/tex]  = 121.5  2.45 10⁻³

      [tex]v_{y \ max}[/tex]  = 0.298 m / s

For point x = 75 cm = 0.750 m

 

We seek the value of

      kx = 8.38  0.750

      kx = 6.285 = 2π

therefore this point is also an antinode and the results do not change.

In conclusion, using the characteristics of standing waves, we can find the results for the questions about the speed of the rope are:

  • The transverse oscillatory velocity is:  [tex]v_y[/tex]  = 0.298 m / s

Learn more here:  https://brainly.com/question/12536719

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