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.

The equation of transverse wave on a rope is:

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

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:

\(v = f\lambda \)

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 \(625\;{\rm{cm}}{{\rm{s}}^{{\rm{ - 1}}}}\).