Question

When a sound wave travels directly toward a hard wall, the incoming and reflected waves can combine to produce a standing wave. There is an antinode right at the wall, just as at the end of a closed tube, so the sound near the wall is loud. You are standing beside a brick wall listening to a 80Hz tone from a distant loudspeaker.
How far from the wall must you move to find the first quiet spot? Assume a sound speed of 340 m/s.

Answer 1

Answer:

L= 1.06 m

Explanation:

frequency of a wave in a pipe closed at one end is given by

$f= \frac{v}{4L}$

L= length of the pipe

v= velocity of wave

therefore, the length of the pipe

$L= \frac{v}{4f}$

$L= \frac{340}{4\times80}$

L= 1.06 m

L= 1.06 m far from the wall must you move to find the first quiet spot.

Answer 2

The distance from the wall(i.e the pipe length) you need to move to find the first quiet spot is 1.06 m.

What is the frequency of a sound wave in a closed pipe?

The smallest standing wave pattern, i.e the first fundamental frequency or harmonic for a closed pipe usually consists of one node and an antinode. When the wavelength of a closed pipe of length L is 4 L, the basic standing wave is formed.

The frequency of a sound wave for a closed pipe at one end may thus be represented using the following formula:

$\mathbf{f = \dfrac{v}{4L}}$

where;

• L =  pipe length = ???
• v = velocity of the wave = 340 m/s
• f = frequency = 80 Hz

Thus, the length of the pipe can be computed as:
$\mathbf{L = \dfrac{340 \ m/s}{4\times 80 \ Hz}}$

L = 1.06 m

Learn more about the frequency of sound waves here:
https://brainly.com/question/15666150