Chapter 14, Problem 89A

(a)

To determine

The fraction of wavelength of water wave when the crest level is $0.18s$ .

Answer to Problem 89A

The movement that takes place from equilibrium level to the crest represents about $\frac{1}{4}$ of a wavelength. The fraction of wavelength is $0.25$ .

Explanation of Solution

Given:s Time of the water wave: $0.18s$ .

Introduction:

The distancebetween the position of equilibrium and the rest is $\frac{1}{4}$ of the wavelength.

The distance is calculated from the equilibrium position.

The movement that takes place from equilibrium level to the crest represents about $\frac{1}{4}$ of a wavelength.

The fraction of wavelength of water wave $=\frac{1}{4}$

  $=0.25$

Conclusion:

Hence, the fraction of wavelength of water wave is $0.25$ .

(b)

To determine

The period of the water wave when the crest level is $0.18s$ .

Answer to Problem 89A

The period of the water wave is $0.72s$ .

Explanation of Solution

Given:

Time of the water wave: $0.18s$ .

Formula used:

To calculate the time periodformula used is,

  $T=4t$

Where,

T: Time

Calculation:

To find the time period of the water wave,

t is given as $0.18s$

  $T=4t$

  $T=4\times 0.18$

  $T=0.72s$

Conclusion:

Hence the time period of the water wave is $0.72s$ .

(c)

To determine

The frequency of water wave when the crest level is $0.72s$ .

Answer to Problem 89A

The frequency of water wave is $1.39$ $\text{Hz}$

Explanation of Solution

Given:

Time of the water wave: $0.18\text{s}$ .

Formula used:

The frequency is the reciprocal of the Time period,

Formula is written as,

  $f=\frac{1}{T}$

Where,

  $f$ : Frequency

  $T$ : Time

Calculation:

As frequency is the reciprocal of Time Period.

  $f=\frac{1}{T}$

  $f=\frac{1}{0.72\text{s}}$

  $f$ = $1.39$

  $\text{Hz}$

Conclusion: The frequency of water wave is $1.39$

  $\text{Hz}$ .