Question

A fellow student with a mathematical bent tells you that the wave function of a traveling wave on a thin rope is\(y\left( {x,t} \right) = 2.30\,mm\,cos\left[ {\left( {6.98\,{{rad} \mathord{\left/{\vphantom {{rad} m}} \right.

\nulldelimiterspace} m}} \right)x + \left( {742\,{{rad} \mathord{\left/{\vphantom {{rad} s}} \right.

\nulldelimiterspace} s}} \right)t} \right]\). Being more practical, you measure the rope to have a length of \(1.35 m\) and a mass of\(0.00338 kg\). You are then asked to determine the following: (a) amplitude; (b) frequency; (c) wavelength; (d) wave speed; (e) direction the wave is traveling; (f) tension in the rope; (g) average power transmitted by the wave.

Short answer

(a) \(A = 2.30\,mm\)

Step-by-step solution

Given data

\(y\left( {x,t} \right) = 2.30\,mm\,cos\left[ {\left( {6.98\,{{rad} \mathord{\left/

{\vphantom {{rad} m}} \right.

\kern-\nulldelimiterspace} m}} \right)x + \left( {742\,{{rad} \mathord{\left/

{\vphantom {{rad} s}} \right.

\kern-\nulldelimiterspace} s}} \right)t} \right]\)

Concept/ Formula used

Amplitude refers to the greatest displacement or distance that a point on a vibrating body or wave may move relative to its equilibrium location. It is equivalent to the vibration path's half-length.

Calculation for Amplitude  

(a)

Comparison with\(y\left( {x,t} \right) = Acos\left( {kx - \omega t} \right)\)gives\(A = 2.30\,mm,\,k = 6.98\,{{rad} \mathord{\left/

{\vphantom {{rad} m}} \right.

\kern-\nulldelimiterspace} m}\) and \(\omega = 742\,{{rad} \mathord{\left/

{\vphantom {{rad} s}} \right.

\kern-\nulldelimiterspace} s}\).

So, Amplitude \(A = 2.30\,mm\)