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Chapter 12: Q61P (page 328)

(II) A wave on the ocean surface with wavelength 44 m travels east at a speed of 18 m/s relative to the ocean floor. If, on this stretch of ocean, a powerboat is moving at 14 m/s (relative to the ocean floor), how often does the boat encounter a wave crest, if the boat is traveling (a) west, and (b) east?

Short Answer

Expert verified

The time period is (a) \(1.4\;{\rm{s}}\) and (b) \(12.5\;{\rm{s}}\).

Step by step solution

01

Understanding the Doppler formula

In the problems, related to relative speed, speed of the sound and frequency the formula of Doppler shift is applied. In estimating the time at which boat encounters a wave crest, use the formula of time period with frequency.

02

Given Data

The wavelength is\(\lambda = 44\;{\rm{m}}\).

The speed of the wave is\({v_1} = 18\;{\rm{m/s}}\).

The speed of the powerboat is \({v_2} = 14\;{\rm{m/s}}\).

03

Evaluating the frequency of the wave

Part (a)

The relation of frequency is given by,

\(\begin{aligned}{l}f &= \frac{{{v_1}}}{\lambda }\\f &= \left( {\frac{{18\;{\rm{m/s}}}}{{44\;{\rm{m}}}}} \right)\\f &= 0.40\;{\rm{Hz}}\end{aligned}\)

The relation from Doppler shift formula is given by,

\({f_0} = f\left( {1 + \frac{{{v_2}}}{{{v_{\rm{1}}}}}} \right)\)

On plugging the values in the above relation.

\(\begin{aligned}{l}{f_{\rm{0}}} &= \left( {0.40\;{\rm{Hz}}} \right)\left( {1 + \frac{{14\;{\rm{m/s}}}}{{18\;{\rm{m/s}}}}} \right)\\{f_{\rm{0}}} &= 0.71\;{\rm{Hz}}\end{aligned}\)

04

Evaluating the time period

The relation of time period is given by,

\({T_1} = \frac{1}{{{f_0}}}\)

On plugging the values in the above relation.

\(\begin{aligned}{l}{T_1} &= \left( {\frac{1}{{0.71\;{\rm{Hz}}}}} \right)\\{T_1} &= 1.4\;{\rm{s}}\end{aligned}\)

Thus, \({T_1} = 1.4\;{\rm{s}}\) is the required time.

05

Evaluating the frequency of the moving observer

Part (b)

The relation from Doppler shift formula is given by,

\({f_0}' = f\left( {1 - \frac{{{v_2}}}{{{v_{\rm{1}}}}}} \right)\)

On plugging the values in the above relation.

\(\begin{aligned}{l}{f_{\rm{0}}} &= \left( {0.40\;{\rm{Hz}}} \right)\left( {1 - \frac{{14\;{\rm{m/s}}}}{{18\;{\rm{m/s}}}}} \right)\\{f_{\rm{0}}} &= 0.08\;{\rm{Hz}}\end{aligned}\)

06

Evaluating the time period

The relation of time period is given by,

\({T_2} = \frac{1}{{{f_0}'}}\)

On plugging the values in the above relation.

\(\begin{aligned}{l}{T_2} &= \left( {\frac{1}{{0.08\;{\rm{Hz}}}}} \right)\\{T_2} &= 12.5\;{\rm{s}}\end{aligned}\)

Thus, \({T_2} = 12.5\;{\rm{s}}\) is the required time.

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