Q48P - University Physics with Modern Physics

Question

A transverse wave on a rope is given by

 

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

 

(a) Find the amplitude, period, frequency, wavelength, and speed of propagation. 

(b) Sketch the shape of the rope at these values of \(t\): \(0,0.0005\;s,0.0010\;s\).

(c) Is the wave travelling in the \( + x\;{\rm{or}} - x\)-direction?

(d) The mass per unit length of the rope is \(0.0500\;{\rm{kg/m}}\). Find the tension.

(e) Find the average power of this wave.

 

Step-by-Step Solution

Answer

Thus, (a) the amplitude is \(0.750\;{\rm{cm}}\), period is \(0.008\;{\rm{s}}\), frequency is \(125\;{\rm{Hz}}\), wavelength is \(5\;{\rm{cm}}\) and speed of propagation is \(625\;{\rm{cm}}{{\rm{s}}^{{\rm{ - 1}}}}\).

 

Step 1: (a) Given in the question

The equation of transverse wave on a rope is:

 

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

Step 2: Explanation of 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: 

\(T = \frac{{2\pi }}{\omega }\)

 

Formula of frequency is:

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

 

Formula of wavelength is:

\(\lambda  = \frac{{2\pi }}{k}\)

 

Formula of speed of propagation:

\(v = f\lambda \)

Step 3: Calculate amplitude, period, frequency, wavelength and speed of propagation.

Compare the given equation with the standard wave equation: 

\(\begin{array}{l}{\rm{y}}\left( {x,t} \right) = \left( {0.750\;{\rm{cm}}} \right)\cos \pi \left[ {\left( {0.400\;{\rm{c}}{{\rm{m}}^{ - 1}}} \right)x + \left( {250\;{{\rm{s}}^{ - 1}}} \right)t} \right]\\{\rm{y}}\left( {x,t} \right) = \left( {0.750\;{\rm{cm}}} \right)\cos 0.4\pi  + 250\pi t\end{array}\)

 

Then it gives:

Amplitude is \(0.750\;{\rm{cm}}\) and the wave number is \(0.4\pi \;{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}\). 

Angular frequency is \(250\pi \) which is written as:

\(\begin{array}{c}\omega  = 250\pi \\\frac{{2\pi }}{T} = 250\pi \\T = \frac{{2\pi }}{{250}}\\ = 0.008\;{\rm{s}}\end{array}\)

Thus, the period is \(0.008\;{\rm{s}}\).

Now, frequency is obtained by using following formula. 

\(\begin{array}{c}f = \frac{1}{T}\\ = \frac{1}{{0.008\;{\rm{s}}}}\\ = 125\;{\rm{Hz}}\end{array}\)

 

Use the formula to calculate wavelength,

\(\begin{array}{c}\lambda  = \frac{{2\pi }}{k}\\ = \frac{{2\pi }}{{0.4\pi \;{\rm{c}}{{\rm{m}}^{{\rm{ - 1}}}}}}\\ = 5\;{\rm{cm}}\end{array}\)

 

Use the formula to calculate speed of wave propagation,

\(\begin{array}{c}v = f\lambda \\ = 125\;{{\rm{s}}^{{\rm{ - 1}}}} \times 5\;{\rm{cm}}\\ = 625\;{\rm{cm}}{{\rm{s}}^{{\rm{ - 1}}}}\end{array}\)

Hence, the amplitude is \(0.750\;{\rm{cm}}\), period is \(0.008\;{\rm{s}}\), frequency is \(125\;{\rm{Hz}}\), wavelength is \(5\;{\rm{cm}}\) and speed of propagation is \(625\;{\rm{cm}}{{\rm{s}}^{{\rm{ - 1}}}}\).