Chapter 12, Problem 67GP

(a)

To determine

To Find:The frequencies that can be heard by visitor’s ear placed near the ends of the pipes.

Answer to Problem 67GP

Frequencies that can be heard by visitor’s ear placed near the ends of each pipe are calculated:

  $f_1=57.16\text{ Hz}$

  $f_2=68.6\text{ Hz}$

  $f_3=85.75\text{ Hz}$

  $f_4=114.33\text{ Hz}$

  $f_5=171.5\text{ Hz}$

Explanation of Solution

Given:

Pipe length $\left(L_1\right) =3\text{ m},$

Pipe length $\left(L_2\right) =2.5\text{ m},$

Pipe length $\left(L_3\right) =2\text{ m},$

Pipe length $\left(L_4\right) =1.5\text{ m},$

Pipe length $\left(L_5\right) =1\text{ m}$

Formula used:

  $f=\frac{c}{2L}\text{ Hz}$

Here $f$ is the frequency,

  $c$ is speed of sound,

  $l$ is the length of pipe.

Calculation:

The pipe is opened at both the ends,

Therefore, the frequency is calculated by the following equation,

  $f=\frac{c}{2L}\text{ Hz}$

Where,

  $c$ = Speed of sound $=343\text{ m/s}$ .

Substitute the values in the given equation, frequency,

For the pipe length $L_1=3\text{ m}$

  $f_1=\frac{343\text{ m/s}}{2\times 3\text{ m}}$

  $f_1=57.16\text{ Hz}$

For the pipe length $L_2=\text{ 2}\text{.5 m}$

  $f_2=\frac{343\text{ m/s}}{2\times 2.5\text{ m}}$

  $f_2=68.6\text{ Hz}$

For the pipe length $L_3=\text{ 2 m}$

  $f_3=\frac{343\text{ m/s}}{2\times 2\text{ m}}$

  $f_3=85.75\text{ Hz}$

For the pipe length $L_4 =1.5\text{m}$

  $f_4=\frac{343\text{ m/s}}{2\times 1.5\text{ m}}$

  $f_4=114.33\text{ Hz}$

For the pipe length $L_5=1\text{m}$

  $f_5=\frac{343\text{ m/s}}{2\times 1\text{ m}}$

  $f_5=171.5\text{ Hz}$

Conclusion:

Frequencies that can be heard by visitor’s ear placed near the ends of each pipe are calculated.

(b)

To determine

To Explain: The display works better on noisy day than on the quiet day.

Explanation of Solution

Introduction:

A noise is an unwanted sound andconsidered as unpleasant, loud or disruptive to hear. The noise and sound are indistinguishable as both are caused due to the vibrations through a medium, such as air or water.

From numerous experiments, it can be concluded that the display works better on a noisy day than the quiet day becausethe vibrational waves activate the systems function. The signals can be enhanced by noise and improve the behaviour. 

If the intensity of noise is increased, anoptimal noise level allows people to hear, and feel better. But too much noise is not desirable because that degrades the performance of the display. 

Conclusion: 

The vibrations on a noisy day are more and they can activate the weak signals.