Sound waves of ν=600 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound= 300 m s- 1 )

Question

Sound waves of ν=600 Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound= 300 m s- 1 ) (A) (7/8)m (B) (3/8) m (C) (1/8) m (D) (1/4) m

Tardigrade Admin · Accepted Answer

The wall acts like a rigid boundary and reflects this wave and sends it back towards the open end. At the open an antinode is formed and a node is formed at the wall. The distance between antinode and node is

\frac{λ}{4}

Therefore, if v be the frequency of note emitted then

λ = \frac{v}{v}

⟹ λ = \frac{300}{600} = \frac{1}{2} m

Maximum amplitude is obtained at a distance

= \frac{λ}{4} = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} m

Q. Sound waves of $ν = 600 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound= $300 m s^{- 1}$ )

$\frac{7}{8} m$

$\frac{3}{8} m$

$\frac{1}{8} m$

$\frac{1}{4} m$

Q. Sound waves of ν=600Hz fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound= 300ms−1 )

87​m

83​m

81​m

41​m

Q. Sound waves of $ν = 600 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which all particles will have a maximum amplitude of vibration will be (speed of sound= $300 m s^{- 1}$ )

$\frac{7}{8} m$

$\frac{3}{8} m$

$\frac{1}{8} m$

$\frac{1}{4} m$