Chapter 14, Problem 65A

(a)

To determine

The speed of sonar signal in water.

Answer to Problem 65A

  $150\text{0}\text{m/s}$ .

Explanation of Solution

Given:

Frequency of sonar signal: $1.00\times {10}^6\text{Hz}$

Wavelength of sonar signal: $1.50\text{mm}$

Formula used:

The formula to calculate the speed is given by

  $v=f\times \lambda$

Where,

  $v$ : Speed

  $\lambda$ : Wavelength

  $f$ : Frequency

Units Conversion,

  $1\text{mm}=0.00\text{1}\text{m}$

Here,

  $1.50\text{mm}$

  $=0.015\text{m}$

Calculation:

Substitute the values in the above formula.

  $=\left(1.00\times {10}^6\right)\times \left(0.0015\right)$

  $=1500\text{m/s}$

Conclusion:

Hence, the speed of water wave in a lake is $1500\text{m/s}$ .

(b)

To determine

The period of sonar signal in water.

Answer to Problem 65A

The period of sonar signal in water is $1.00\times {10}^{-6}$ .

Explanation of Solution

Given:

Frequency of sonar signal: $1.00\times {10}^6\text{Hz}$

Wavelength of sonar signal: $1.50\text{mm}$

Formula used:

The formula to calculate the period is given by

  $T=\frac{1}{f}$

Where,

  $f$ : Frequency

  $T$ : Time Period

Calculation:

The formula to calculate the period is given by,.

  $T=\frac{1}{f}$

Substitute the values in the above formula.

  $T=\frac{1}{1.00\times {10}^6}$

  $T=1.00\times {10}^{-6}s$

Conclusion:

Hence, the period of sonar signal in water is $1.00\times {10}^{-6}\text{s}$ .