Question

A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]).

Part A Find the amplitude.

Part B Find the period

Part C Find the frequency

Part D Find the wavelength

Part E Find the speed of propagation.

Part F Is the wave traveling in the +x- or − x-direction?

Answer 1

Concepts and reason

The concepts required to solve this problem is the wave-equations for a wave travelling in positive and negative x directions in term of amplitude, wavelength, angular frequency, time and position.

First write the general wave equation in terms of amplitude, wavelength, frequency, time and position.

Then, compare the general wave equation with given wave equation to find the amplitude, period, frequency, wavelength, speed of propagation, and direction of propagation of wave.

Fundamentals

The wave equation for wave travelling in positive x direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx - \omega t} \right)$

Here, y is the position of y-coordinate of the wave at time t. x is the position of x-coordinate of the wave at time t. A is the amplitude of the wave. k is the wave-vector of the wave, $\omega$ is the angular frequency of the wave.

The wave equation for wave travelling in negative x direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)$

The relation between wavevector and wavelength is given as follows:

$k = \frac{{2\pi }}{\lambda }$

Here, $\lambda$ is the wavelength of the wave.

The relation between angular frequency and frequency is given as follows:

$\omega = 2\pi f$

Here, f is the wavelength of the wave.

The speed of the propagation of the wave in terms of wavevector and angular frequency is given as follows:

$v = \frac{\omega }{k}$

Here, v is the velocity of the wave.

The relation between the time-period of the wave and frequency of the wave is given as follows:

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

Here, T is the time-period of the wave.

(Part A)

The wave-equation of wave travelling in negative x-direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)$

The wave-equation in problem is given as follows:

$y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)$

Compare the wave-equation of wave travelling in negative x-direction and wave-equation in problem and determine the amplitude of the wave.

Therefore, the amplitude of the wave is 0.750 cm.

Part B

The wave-equation of wave travelling in negative x-direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)$

The wave-equation in problem is given as follows:

$y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)$

Compare the wave-equation of wave travelling in negative x-direction and wave-equation in problem and determine the angular frequency of the wave.

Therefore, the angular frequency of the wave is $250\pi {\rm{ }}{{\rm{s}}^{ - 1}}$ .

The relation between the angular frequency and time-period of the wave is given as follows:

$T = \frac{{2\pi }}{\omega }$

Substitute $250\pi {\rm{ }}{{\rm{s}}^{ - 1}}$ for $\omega$ in the above-mentioned equation.

$\begin{array}{c}\\T = \frac{{2\pi }}{{\left( {250\pi {\rm{ }}{{\rm{s}}^{ - 1}}} \right)}}\\\\ = 0.008{\rm{ s}}\\\end{array}$

Therefore, time-period of the wave is 0.008 s.

Part C

The relation between frequency and time-period is given as follows:

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

Substitute 0.008 s for T in the above-mentioned equation.

$\begin{array}{c}\\f = \frac{1}{{\left( {0.008{\rm{ s}}} \right)}}\\\\ = \left( {125{\rm{ }}{{\rm{s}}^{ - 1}}} \right)\left( {\frac{{{\rm{Hz}}}}{{1{\rm{ }}{{\rm{s}}^{ - 1}}}}} \right)\\\\ = 125{\rm{ Hz}}\\\end{array}$

Therefore, the frequency of the wave is 125 Hz.

Part D

The wave-equation of wave travelling in negative x-direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)$

The wave-equation in problem is given as follows:

$y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)$

Compare the wave-equation of wave travelling in negative x-direction and wave-equation in problem and determine the wavevector of the wave.

Therefore, the wavevector of the wave is $0.400\pi {\rm{ c}}{{\rm{m}}^{ - 1}}$ .

The relation between wavelength and wavevector is given as follows:

$\lambda = \frac{{2\pi }}{k}$

Substitute $0.400\pi {\rm{ c}}{{\rm{m}}^{ - 1}}$ for k in the above-mentioned equation.

$\begin{array}{c}\\\lambda = \frac{{2\pi }}{{\left( {0.400\pi {\rm{ c}}{{\rm{m}}^{ - 1}}} \right)}}\\\\ = 5{\rm{ cm}}\\\\ = 5{\rm{ cm}}\\\end{array}$

Therefore, the wavelength of the wave is 5 cm.

Part E

The relation between wavevector and angular frequency is given as follows:

$v = \frac{\omega }{k}$

Substitute $0.400\pi {\rm{ c}}{{\rm{m}}^{ - 1}}$ for k and $250\pi {\rm{ }}{{\rm{s}}^{ - 1}}$ for $\omega$ in the above-mentioned equation.

$\begin{array}{c}\\v = \frac{\omega }{k}\\\\ = \frac{{250\pi {\rm{ }}{{\rm{s}}^{ - 1}}}}{{0.400\pi {\rm{ c}}{{\rm{m}}^{01}}}}\\\\ = 625{\rm{ cm/s}}\\\end{array}$

Therefore, the speed of the wave is $625{\rm{ cm/s}}$ .

Part F

The wave equation for wave travelling in positive x direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx - \omega t} \right)$

The wave equation for wave travelling in negative x direction is given as follows:

$y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)$

The wave-equation in problem is given as follows:

$y\left( {x,t} \right) = \left( {0.750{\rm{ cm}}} \right)\cos \left( {\pi \left( {0.400{\rm{ c}}{{\rm{m}}^{ - 1}}} \right)x + \pi \left( {250{\rm{ }}{{\rm{s}}^{ - 1}}} \right)t} \right)$

Comparing the given wave equation with the wave equations travelling in positive and negative x-directions. It is determined that the wave is travelling in negative x-direction.

Therefore, the direction of propagation of wave is in negative x-direction.

Ans: Part A

The amplitude of the wave is 0.750 cm.

Part B

Time-period of the wave is 0.008 s.

Part C

Frequency of the wave is 125 Hz.

Part D

The wavelength of the wave is 5 cm.

Part E

The speed of the wave is ${\bf{625 cm/s}}$ .

Part F

The direction of propagation of wave is in negative x-direction.