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Chapter 2: Q8E (page 497)

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.

Short Answer

Expert verified

(a) \(6.50\,mm\)

Step by step solution

01

Given Data

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

02

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.

03

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 \(6.50\,mm\)

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Most popular questions from this chapter

(a) A sound source producing 1.00-kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare to that in part (a)? Explain on physical grounds why the two answers differ.

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