Q10E - University Physics with Modern Physics

Question

A water wave traveling in a straight line on a lake is described by the equation

               \(y\left( {x,t} \right) = \left( {2.75\,cm} \right)cos\left( {0.410\,{{rad} \mathord{\left/

 {\vphantom {{rad} {cm}}} \right.

 \kern-\nulldelimiterspace} {cm}}\,x + 6.20{{rad} \mathord{\left/

 {\vphantom {{rad} s}} \right.

 \\,t} \right)\) 

Where \(y\) is the displacement perpendicular to the undisturbed surface of the lake. (a) How much time does it take for one complete wave pattern to go past a fisherman in a boat at anchor, and what horizontal distance does the wave crest travel in that time? (b) What are the wave number and the number of waves per second that pass the fisherman? (c) How fast does a wave crest travel past the fisherman, and what is the maximum speed of his cork floater as the wave causes it to bob up and down?

Step-by-Step Solution

Answer

(a) \(T = 1.013\,s\) 

 \(\lambda  = 0.153\,m\) 

Step 1: Given data

\(y\left( {x,t} \right) = \left( {2.75\,cm} \right)cos\left( {0.410\,{{rad} \mathord{\left/

 {\vphantom {{rad} {cm}}} \right.

 \kern-\nulldelimiterspace} {cm}}\,x + 6.20{{rad} \mathord{\left/

 {\vphantom {{rad} s}} \right.

 \\,t} \right)\)

Step 2: Concept/ Formula used

Time period is defined as the time taken by the wave to complete one vibrational cycle.

Formula: 

\(T = \frac{{2\pi }}{\omega }\) 

Where, \(\omega \)is angular frequency in \({{rad} \mathord{\left/

 {\vphantom {{rad} s}} \right.

 \\nulldelimiterspace} s}\) 

Step 3: Calculation of Time period

(a) 

By comparison \(y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)\) with given wave equation.

\(\begin{aligned}{l}A = 2.75\,cm\\k = 0.410\,{{rad} \mathord{\left/

 {\vphantom {{rad} {cm}}} \right.

 \\nulldelimiterspace} {cm}}\\\omega  = 6.20\,{{rad} \mathord{\left/

 {\vphantom {{rad} s}} \right.

 \\nulldelimiterspace} s}\end{aligned}\) 

\(\begin{aligned}{c}T = \frac{{2\pi }}{\omega }\\ = \frac{{2\pi \,rad}}{{6.20\,{{rad} \mathord{\left/

 {\vphantom {{rad} s}} \right.

 \\nulldelimiterspace} s}}}\\ = 1.013\,s\end{aligned}\)

Step 4: Calculation for distance travel

Distance travels a wave crest in one cycle is 

\(\begin{aligned}{c}\lambda  = \frac{{2\pi \,}}{k}\\ = \frac{{2\pi \,rad}}{{0.410\,{{rad} \mathord{\left/

 {\vphantom {{rad} {cm}}} \right.

 \kern-\nulldelimiterspace} {cm}}}}\\ = 15.32\,cm = 0.153\,m\end{aligned}\)