Chapter 14, Problem 63A

(a)

To determine

The speed of the water waves.

Answer to Problem 63A

  $2.9\text{m/s}$

Explanation of Solution

Given:

Wavelength of the wave ( $\lambda$ ) is $6.0\text{cm}$ .

Rate of oscillations ( $f$ ) is $4.8\text{oscillations}\text{/s}$ .

Formula used:

The formula to calculate speed of wave is

  $v=\lambda f$

Where,

  $v$ : Velocity

  $\lambda$ : Wavelength

  $f$ : Frequency

Unit Conversion,

  $1\text{cm}=0.01\text{m}$

Calculation:

Substituting the values in the equation:

  $\lambda =\left(0.60\right)\times \left(4.8\right)$

  $=2.9\text{m/s}$

Conclusion:

The speed of the water wave is $=2.9\text{m/s}$ .

(b)

To determine

The time period of wave.

Answer to Problem 63A

  $0.21\text{m/s}$

Explanation of Solution

Given:

Rate of oscillations is $4.8\text{s}$

Formula used:

The period formula is:

  $T=\frac{1}{f}$

Where,

  $f$ : Frequency

T: Period

Calculation:

To calculate the period, substitute the values in the above formula

  $T=\frac{1}{f}$

  $T=\frac{1}{\text{4}\text{.8}\text{s}}$

  $T$ = $\text{0}\text{.21}\text{s}$

Conclusion:

Hence, the period of wave is $0.2\text{1}\text{s}$ .