Question

A transverse harmonic wave travels on a rope according to the following expression:

Answer 1

1) Amplitude: 0.14 m

2) Frequency: 2.8 Hz

3) Wavelength: 3.0 m

4) Speed: 8.4 m/s

5) Tension: 7.3 N

6) Velocity at x=3.4 m and t=0.48 s: -2.47 m/s

7) Acceleration: $-3.16 m/s^2$

8) Average speed of the rope: 0.25 m/s

9) The wave is travelling in the +x direction

10) The period would increase

Explanation:

1)

The equation that describes the displacement of a wave is in the form

$y(x,t)=A sin (kx-\omega t)$ (1)

where

A is the amplitude

$k=\frac{2\pi}{\lambda}$ is the wave number, with $\lambda$ being the wavelength

x is the position

$\omega$ is the angular frequency

t is the time

The equation for the wave in this problem is

$y(x,t) = 0.14sin(2.1x + 17.7t)$ (2)

By comparing (1) and (2), we see that the amplitude is A = 0.14 m.

2)

The angular frequency of the wave can be written as

$\omega = 2\pi f$ (3)

where f is the frequency. For the wave in the problem, by comparing (1) and (2),

$\omega=17.7 rad/s$

Re-arranging the equation (3), we can find the frequency:

$f=\frac{\omega}{2\pi}=\frac{17.7}{2\pi}=2.8 Hz$

3)

The wave number of the wave is related to the frequency by

$k=\frac{2\pi}{\lambda}$ (4)

where $\lambda$ is the wavelength.

By comparison of (1) and (2), we know that

$k=2.1 m^{-1}$

And solving the equation (4) for the wavelength, we find:

$\lambda=\frac{2\pi}{k}=\frac{2\pi}{2.1}=3.0 m$

4)

The speed of a wave is given by the following equation, also known as wave equation:

$v=f\lambda$

where

f is the frequency

$\lambda$ is the wavelength

For this wave, we have

f = 2.8 Hz

$\lambda=3.0 m$

Substituting,

$v=(2.8)(3.0)=8.4 m/s$

5)

The speed of the standing waves in a string is given by the equation

$v=\sqrt{\frac{T}{\mu}}$

where

T is the tension in the string

$\mu$ is the linear mass density of the string

For the string in this problem, we know:

v = 8.4 m/s is the speed of the wave

$\mu=0.104 kg/m$ is the linear mass density

Solving for T, we find the tension in the string:

$T=\mu v^2 = (0.104)(8.4)^2=7.3 N$

6)

The velocity of the rope can be found calculating the derivative of the displacement, therefore it is:

$y'(x,t)=(0.14)(17.7) cos(2.1x+17.7t)=2.48 cos (2.1x+17.7t)$

Here we want to know the velocity at the following conditions:

x = 3.4 m

t = 0.48 s

Substituting into the equation of y', we find:

$y'=2.48 cos (2.1(3.4)+17.7(0.48))=-2.47 m/s$

7)

The acceleration of the rope can be found calculating the derivative of the velocity, therefore it is:

$y''(x,t)=-(2.48)(17.7) sin(2.1x+17.7t)=-43.9 sin (2.1x+17.7t)$

Here we want to know the acceleration at the following conditions:

x = 3.4 m

t = 0.48 s

Substituting into the equation of y'', we find:

$y''=-43.9 sin (2.1(3.4)+17.7(0.48))=-3.16 m/s^2$

8)

To find the average speed of the rope, we have to divide the total distance covered in one oscillation by the period of the wave.

The distance covered by the rope during one oscillation is equal to 4 times the amplitude:

$d=4A=4(0.14)=0.56 m$

While the period can be found from the angular frequency:

$T=\frac{2\pi}{f}=\frac{2\pi}{2.8}=2.24 s$

Therefore, the average speed of the wave is:

$v=\frac{d}{T}=\frac{0.56}{2.24}=0.25 m/s$

9)

The equation of a wave can be written in two ways:

• $y(x,t)=A sin (kx-\omega t)$: this is the equation of a wave travelling in the +x direction

• $y(x,t)=A sin(kx+\omega t)$: this is the equation of a wave travelling in the -x direction

For the wave in this problem, the equation is written as

$y(x,t)=A sin (kx-\omega t)$

This means that it is a wave travelling in the +x direction.

10)

As we said, the wavelength is related to the speed of the wave and to the frequency by

$v=f \lambda$

The frequency can be written as a function of the period,

$f=\frac{1}{T}$

So the previous equation becomes

$v=\frac{\lambda}{T}$

or

$T=\frac{\lambda}{v}$

The speed of the wave (v) depends only on the properties of the string, therefore it does not change; so, if we increases the wavelength, the period would increase.

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