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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.

Short Answer

Expert verified

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 by step solution

01

(a) Given in the question

The equation of transverse wave on a rope is:

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

02

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:

\(v = f\lambda \)

03

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

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Most popular questions from this chapter

A heavy rope 6.00 m long and weighing 29.4 N is attached at one end to a ceiling and hangs vertically. A 0.500-kg mass is suspended from the lower end of the rope. What is the speed of transverse waves on the rope at the (a) bottom of the rope, (b) middle of the rope, and (c) top of the rope? (d) Is the tension in the middle of the rope the average of the tensions at the top and bottom of the rope? Is the wave speed at the middle of the rope the average of the wave speeds at the top and bottom? Explain.

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