Question

A simple harmonic oscillator at the point x = 0 gener ates a wave on a rope. The oscillator operates at a frequency of 40.0 Hz and with an amplitude of 3.00 cm. The rope has a linear mass density of 50.0 g/m and is stretched with a tension of 5.00 N. (a) Determine the speed of the wave. (b) Find the wavelength. (c) Write the wave function y(x, t) for the wave. Assume that the oscillator has its maximum upward displacement at time t = 0. (d) Find the maximum transverse acceleration of points on the rope. (e) In the discussion of transverse waves in this chapter, the force of gravity was ignored. Is that a reasonable approximation for this wave? Explain.

Short answer

The speed of the wave is v = 10.0m/s, the wavelength of the wave is λ = 0.250 m,the wave function of the wave is $y\left(x,t\right)=\left(0.030\right)\cos \left[80.0\pi \left(\frac{x}{10.0}\mp t\right)\right]$, the maximum transverse acceleration is role="math" localid="1664342285483" $a_{y.max}=1895m/s^2$and yes, this is a reasonable approximation.

Step-by-step solution

Determination of the formula of Mechanical Waves

The wave function y(x, t) of a sinusoidal wavedescribes the displacement of individual particles in the medium:

$y\left(x,t\right)=A\cos \left[80.0\pi \left(\frac{x}{v}\mp t\right)\right]$--(1)

Where A, f and v are constants.

The minus sign is used when the wave is traveling in +X-direction and the plus sign is used when the wave is traveling in -X-direction.

The wave speed on a string in terms of the tension F and the linear mass density is:

$v=\sqrt{\frac{T}{\mu }}$--(2)

The wave speed of a periodic wave in terms of the frequency f and the wavelength is:

$v=f\times \lambda$--(3)

The frequency of the oscillator is f = 40.0 Hz, the amplitude is A = 3.00 cm = 0.030 m, the linear mass density of the rope is μ = 50.0 g/m = 0.050 kg/m and the tension in the rope is T = 5.00 N.

Application of the formula of Mechanical Waves

(a) Put in the values for T and into equation (2):

$$\begin{gathered} v=\sqrt{\frac{5.00N}{0.050kg/m}} \\ =10.0m/s \\ v=10.0m/s \end{gathered}$$

(b) Substitute for v and f into equation (3) and solve for λt:

$$\begin{gathered} 10m/s=\left(40.0s^{-1}\right)\times \lambda \\ \lambda =0.25m \end{gathered}$$

(c) Since this is a sinusoidal wave starting from x = 0 at t = 0, substitute for f, v andA into equation (1):

$y\left(x,t\right)=\left(0.030\right)\cos \left[80.0\pi \left(\frac{x}{10.0}\mp t\right)\right]$

(d) The transverse acceleration of a sinusoidal wave is the second derivative of the transverse displacement y(x,t) with respect to time t.

So, the second derivative of this expression for y(x, t):

role="math" localid="1664342524957" $$\begin{gathered} a_y=\frac{{\partial }^{2}y\left(x,t\right)}{\partial t^2}-\left[\right(0.030)\times {(0.80\pi )}^2\cos \left[80.0\pi \left(\frac{x}{10.0}\mp t\right)\right] \\ =192{\pi }^2\cos \left[80.0\pi \left(\frac{x}{10.0}\mp t\right)\right] \end{gathered}$$

The maximum acceleration occurs when the value of the cosine function equals 1, so the maximum acceleration of this wave is:

$$\begin{gathered} a_{y.max}=192{\pi }^2 \\ {a}_{y.max}=1895m/s^2 \end{gathered}$$

Determination of a reasonable approximation

(e) Yes, this a reasonable approximation because$a_{y.max}$ is much larger than g;

Which means that the force caused by the perturbation on the rope is much larger than the weight of the rope.


Therefore, the speed of the wave is v = 10.0m/s, the wavelength of the wave is λ = 0.250 m,the wave function of the wave is $y\left(x,t\right)=\left(0.030\right)\cos \left[80.0\pi \left(\frac{x}{10.0}\mp t\right)\right]$, the maximum transverse acceleration is$a_{y.max}=1895m/s^2$ and yes, this is a reasonable approximation.