Chapter 14, Problem 67A

(a)

To determine

To Find: The frequency of the wave.

Answer to Problem 67A

The frequency of the sound is $550\text{Hz}$ .

Explanation of Solution

Given:

Wavelength of the signal is $\text{0}\text{.6}\text{m}$ .

Velocity of sound wave is $\text{330 m/s}$ .

Time period is $\text{0}\text{.5 s}$

Formula Used:

Frequency of the wave is

  $f=\frac{v}{\lambda }$

…… (1)

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

Calculation:

Substitute $330\text{m/s}$ for $v$ , $\text{0}\text{.5 s}$ for $t$ in equation (1)

  $\begin{array}{l}f=\frac{v}{\lambda }\\ f=\frac{330}{0.6}\\ f=550\text{Hz}\end{array}$

Conclusion:

Hence, the frequency of the sound is $550\text{Hz}$ .

(b)

To determine

To Find: The number of complete waves emitted in the given period.

Answer to Problem 67A

275

Explanation of Solution

Given:

Wavelength of the signal is $\text{0}\text{.6}\text{m}$ .

Velocity of $\text{330 m/s}$ .

Time period is $\text{0}\text{.5 s}$ .

Formula:

Number of complete waves that is emitted in time interval $T$ is

  $n=fT$

…… (1)

Here, $n$ is the number of complete wave, $f$ is the frequency and $T$ is the time period.

Calculation:

Substitute $550\text{Hz}$ for $f$ , $\text{0}\text{.5 s}$ for $T$ in equation (1)

  $\begin{array}{l}n=fT\\ n=550\times 0.5\\ n=275\end{array}$

The number complete wave is $\text{275}$ .

Conclusion:

Hence, the number complete wave is $\text{275}$ .

(c)

To determine

To Find: The distance after $\text{0}\text{.5 s}$ .

Answer to Problem 67A

The distance travel by the wave is $\text{165}\text{m}$ .

Explanation of Solution

Given:

Wavelength of the signal is $\text{0}\text{.6}\text{m}$ .

Velocity of sound wave is $\text{330 m/s}$ .

Time period is $\text{0}\text{.5 s}$ .

Formula:

Distance travelled by the wave is

  $d=vT$

Here, $V$ and $T$ is the speed and time period respectively.

Calculation:

Substitute $330\text{m/s}$ for v , $\text{0}\text{.5 s}$ for $T$ in equation (1),

  $\begin{array}{l}d=vT\\ d=330\times 0.5\\ d=165\text{m}\end{array}$

Conclusion:

Hence, the distance travel by the wave is $\text{165}\text{m}$ .