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Fundamentals Of Physics
Found in: Page 472

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Short Answer

A sinusoidal wave moving along a string is shown twice in Figure 16-33, as crest A travels in the positive direction of an x axis by distance d = 6.0 cm in 4.0 ms . The tick marks along the axis are separated by 10 cm ; height H = 6.00mm If the wave equation is of the form, y(x,t)=ymsin(kx+ωt)

(a) What is ym ,

(b) What is k ,

(c) What is ω , and

(d) What is the correct choice of sign in front of ω?

a) The maximum amplitude is 3.00×10-3m.

b) The wave vector is16.0 m-1 .

c) The angular frequency is 2.4×102 rad/s.

d) The correct choice of sign in front of ω is negative.

See the step by step solution

Step by Step Solution

Step 1: The given data

  • The distance travelled by the crest,d = 6.0 cm or 0.06 m .
  • The time required for this displacement,t=4.0 ms or 4.0×10-3s .
  • The scale on x -axis, the distance between the tick marks is10 cm .
  • The peak-to-peak distance,H=6.00 mm or 6.00×10-3m .

Step 2: Understanding the concept of wave equation

A sinusoidal wave traveling in positive x-direction is described by a standard equation. We use the equation and the information from the graph to calculate the required quantities.

Formula:

The maximum amplitude of the wave, ym=12×peak to peak distance (i)

The transverse speed of a wave, v=λf=ωk (ii)

The wavenumber of the wave, k=2πλ (iii)

The velocity of a body in motion, v=dt (iv)

Step 3: a) Calculation of the value of ym

It is given that the peak-to-peak distance isH=6.00×10-3m.

The maximum displacement is given using equation (i) and the given values as follows:

ym=12×6.00×10-3=3.00×10-3m

Hence, the value of maximum amplitude is 3.00×10-3m .

Step 4: b) Calculation of wavevector of the wave

From the graph, we can observe that the graph repeats itself after travelling a distance of 4 tick marks i.e. distance of (4 X 10 cm) = 40 cm

So we get,λ=40.0 cm or 40.0×10-2 m

Now using equation (iii) and given values, the wavevector can be given as:

K=2×3.1440.0×10-2 16.0m-1

Hence, the value of wave vector is16.0m-1 .

Step 5: c) Calculation of angular frequency

Crest A moves distance d in time t in the positive direction of x axis. Thus, the wave velocity using equation (iv) is given as:

v=6.0×10-24.0×10-3 =15 m/s

For a travelling wave, the wave velocity using equation (ii) is given as:

ω=v k =15×16 =240 =2.4×102 rad/s

Hence, the value of angular velocity is2.4×102 rad/s .

Step 6: d) Finding the sign of angular velocity

The sign of ω should be negative as the wave is moving along the positive direction of x axis.

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