Chapter 14, Problem 91A

(a)

To determine

To Describe: The type of wave experienced in floating water.

Explanation of Solution

Introduction:

While floating offshore in a beach, even though the wave pulses travel towards the beach, a person does not move much closer to the beach. The person is floating on a boat.

The type of wave experienced by the person while floating on water is transverse wave. In this case, the displacement of water by the boat is perpendicular to direction of waves.

(b)

To determine

To Explain: The reason of the wave not moving closer to the shore.

Explanation of Solution

Introduction:

While floating offshore in a beach, a person does not move much closer to the beach although the wave pulses travel towards the beach. The person floats on a boat.

As the wave is a transverse wave, the energy of the wave does not move the person closer to the shore.

(c)

To determine

To Evaluate: The period of the waves in $15\text{s}$ when 10 waves passes by.

Answer to Problem 91A

The time period is $1.5\text{s}$ .

Explanation of Solution

Given:

The period of waves is $T=15\text{s}$.

Number of waves is $n=10$.

Formula used:

Time period can be expressed as,

  $t=\frac{T}{n}$….. (1) Here, $T$ and $n$ are the time and wave number.

Calculation:

Substitute $15\text{s}$ for $T$ and $10$ for $n$ in equation (1)

  $\begin{array}{l}t=\frac{15}{10}\\ t=1.5\text{s}\end{array}$

Conclusion:

Hence, the required time period is $1.5\text{s}$ .

(d)

To determine

To Find: The frequency of the waves.

Answer to Problem 91A

The frequency is $f=0.67\text{Hz}$ .

Explanation of Solution

Given:

The time period is $T=1.5\text{s}$.

Formula used:

The frequency is given by

  $f=\frac{1}{T}$.........(1)

Here, $T$ is the time period.

Calculation:

Substitute $1.5\text{s}$ in equation (1),

  $\begin{array}{l}f=\frac{1}{1.5\text{s}}\\ f\text{= 0}\text{.67 Hz}\end{array}$ .

Conclusion:

Hence, the required frequency is $f=0.67\text{Hz}$ .

(e)

To determine

To Find: The velocity of the waves

Answer to Problem 91A

The velocity is $2\text{m/s}$ .

Explanation of Solution

Formula used:

The relation between the velocity and wavelength is given by,

  $v=\lambda f$.........(1)

Here, $\lambda$ is the wavelength, $v$ is the velocity and $f$ is the frequency.

Calculation:

Substituting $3\text{ m}$ in $\lambda$ and $\text{0}\text{.67 Hz}$ in $f$ in equation (1)

  $\begin{array}{c}v=3\times 0.67\\ v=2\text{ m/s}\end{array}$

Conclusion:

Hence, the required velocity is $2\text{m/s}$ .

(f)

To determine

To Describe: The actual wavelength of the waves if the waves are moving at $\text{1}\text{.8 m/s}$ .

Explanation of Solution

Given:

Speed of waves, $v=1.8\text{m/s}$

Frequency of waves, $f=0.67\text{Hz}$

Formula Used:

The relation between the velocity and wavelength is given by,

  $v=\lambda f$.........(1)

Here, $\lambda$ is the wavelength, $v$ is the velocity and $f$ is the frequency.

Calculation:

Substituting $\text{1}\text{.8 m/s}$ in $v$ and $\text{0}\text{.67 s}$ in $f$ in equation (1)

  $\begin{array}{l}1.8=\lambda \times 0.67\\ \lambda =2.68\text{ m}\end{array}$

Conclusion:

Hence, the actual wavelength of the waves is $2.68\text{ m}$ .