Q49P - University Physics with Modern Physics

Question

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels from left to right along a long, horizontal, stretched string with a speed of 36.0 m/s. Take the origin at the left end of the string has its maximum upward displacement.

 

(a) What are the frequency, angular frequency, and wave number of the wave? 

(b) What is the function \(y\left( {x,t} \right)\) that describes the wave?

(c) What is \(y\left( t \right)\)for a particle at the left end of the string?

(d) What is \(y\left( t \right)\) for a particular 1.35 m to right of the origin? 

(e) What is the maximum magnitude of transverse velocity of any particle of the string?

(f) Find the transverse displacement and the transverse velocity of a particle 1.35 m to the right of the origin at time \(t = 0.0625\;s\).

Step-by-Step Solution

Answer

Thus, (a) the frequency is \(20\;{\rm{Hz}}\), angular frequency is \(40\pi \;{{\rm{s}}^{{\rm{ - 1}}}}\) and wave number is \(3.49\;{{\rm{m}}^{{\rm{ - 1}}}}\).

Step 1: (a) Given in the question

Amplitude is \(A = 2.50\;mm\)  

Wavelength is \(\lambda   = 1.80\;{\rm{m}}\).

Speed of propagation is \(v = 36.0\;{\rm{m/s}}\).

Step 2: Standard wave equation

The standard wave equation is: 

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

Here, \(A\) is amplitude, \(\omega \) is angular frequency, \(k\) is wave number.

Formula for period is: 

\(\omega  = 2\pi f\)

 

Formula of frequency is:

\(f = \frac{1}{T}\)

 

Formula of wave number is: 

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

 

Formula of speed of propagation:

\(v = f\lambda \)

Step 3: Calculate frequency, angular frequency, and wave number

Use the formula to calculate frequency, 

\(\begin{array}{c}f = \frac{v}{\lambda }\\ = \frac{{36\;{\rm{m/s}}}}{{1.80\;{\rm{m}}}}\\ = 20\;{{\rm{s}}^{{\rm{ - 1}}}}\\ = 20\;{\rm{Hz}}\end{array}\)

 

Use the formula to calculate angular frequency, 

\(\begin{array}{c}\omega  = 2\pi f\\ = 2\pi  \times 20\;\\ = 40\pi \;{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\)

 

Use the formula to calculate wave number,

\(\begin{array}{c}\lambda  = \frac{{2\pi }}{k}\\1.80\;m = \frac{{2\pi }}{k}\\k = \frac{{2\pi }}{{1.80\;{\rm{m}}}}\\ = 3.49\;{{\rm{m}}^{{\rm{ - 1}}}}\end{array}\)

 

Hence, the frequency is \(20\;{\rm{Hz}}\), angular frequency is \(40\pi \;{{\rm{s}}^{{\rm{ - 1}}}}\) and wave number is \(3.49\;{{\rm{m}}^{{\rm{ - 1}}}}\).