Q8E - University Physics with Modern Physics

Question

A certain transverse wave is described by

\(y\left( {x,t} \right) = \left( {6.50\,mm} \right)cos2\pi \left( {\frac{x}{{28.0\,{\kern 1pt} cm}} - \frac{t}{{0.0360\,s}}} \right)\) 

Determine the wave’s (a) amplitude; (b) wavelength; (c) frequency; (d) speed of propagation; (e) direction of propagation.

Step-by-Step Solution

Answer

(a) \(6.50\,mm\)

Step 1: Given Data

\(y\left( {x,t} \right) = \left( {6.50\,mm} \right)cos2\pi \left( {\frac{x}{{28.0\,{\kern 1pt} cm}} - \frac{t}{{0.0360\,s}}} \right)\)

Step 2: Concept/ Formula used

The largest displacement or distance made by a point on a wave or vibrating body relative to its equilibrium position is its amplitude. It is equal to one-half the length of the vibration path.

Step 3: Calculation for Amplitude

(a) 

Compare given \(y\left( {x,t} \right)\)in the problem to the general form 

\(y\left( {x,t} \right) = A\cos 2\pi \left( {\frac{x}{\lambda } - \frac{t}{T}} \right)\) 

The comparison gives \(A = 6.50\,mm\) 

So, Amplitude is \(6.50\,mm\)