Question

SOUND WAVES OF F=600Hz FALL NORMALLY ON A PERFECTLY REFLECTING WALL . THE SHORTEST DISTANCE AT WHICH ALL PARTICLES WILL HAVE MAX. AMPLITUDE OF VIBRATION IS :
1. 7/8m
2. 3/8m
3. 1/8m
4. 1/4m

Answer 1

Answer on Question #42022, Physics, Mechanics | Kinematics | Dynamics

SOUND WAVES OF $F = 660\mathrm{Hz}$ FALL NORMALLY ON A PERFECTLY REFLECTING WALL. THE SHORTEST DISTANCE AT WHICH ALL PARTICLES WILL HAVE MAX. AMPLITUDE OF VIBRATION IS :

1. $7 / 8\mathrm{m}$

2. $3 / 8\mathrm{m}$

3. $1 / 8\mathrm{m}$

4. $1 / 4\mathrm{m}$

Solution:

Reflected frequency (red) reflects back in-phase, resulting in an increase in amplitude (blue).

The shortest distance at which all particles will have maximum amplitude is on distance of $\lambda / 4$ form the wall:

$$l _ {1} = \frac {\lambda}{4}$$

The wavelength is

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

where $v = 330 \, \text{m/s}$ is velocity of the sound wave, and $f$ is frequency.

Thus,

$$\lambda = \frac {330}{660} = \frac {1}{2} \mathrm {m}$$ $$l _ {1} = \frac {1}{8} \mathrm {m}$$

Answer. 3. $1 / 8 \mathrm{~m}$

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