Parker Pitts

2022-10-03

When a sound wave travels directly toward a hard wal, the incoming and rerlected 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 50 Hz 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.

ordonansexa

Beginner2022-10-04Added 7 answers

To solve this problem, we will be applying an equation that determines the wavelength of the wave:

$\lambda =\frac{v}{f}$

Where: $\lambda $ - wavelength v-wave speed f-frequency

Also, we will be applying the following equation that determines the distance of the first quiet spot:

$d=\frac{\lambda}{4}$

Where: d-distance $\lambda $ - wavelength

First, we will put known values into the equation that determines the wavelength and calculate it as:

$\lambda =\frac{340\frac{m}{s}}{50Hz}$

=6.8m

Next, we will put known values into the equation that determines distance and calculate it:

$d=\frac{6.8m}{4}$

=1.7m

Result:

d=1.7m

$\lambda =\frac{v}{f}$

Where: $\lambda $ - wavelength v-wave speed f-frequency

Also, we will be applying the following equation that determines the distance of the first quiet spot:

$d=\frac{\lambda}{4}$

Where: d-distance $\lambda $ - wavelength

First, we will put known values into the equation that determines the wavelength and calculate it as:

$\lambda =\frac{340\frac{m}{s}}{50Hz}$

=6.8m

Next, we will put known values into the equation that determines distance and calculate it:

$d=\frac{6.8m}{4}$

=1.7m

Result:

d=1.7m

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