MCQ
A sound source of frequency $170 Hz$ is placed near a wall. A man walking from a source towards the wall finds that there is a periodic rise and fall of sound intensity. If the speed of sound in air is $340\, m/s$ the distance (in metres) separating the two adjacent positions of minimum intensity is
  • A
    $0.5$
  • $1$
  • C
    $1.5$
  • D
    $2$

Answer

Correct option: B.
$1$
b
(b) $v = n\lambda $ ==> $\lambda = \frac{v}{n} = \frac{{340}}{{170}}$==> $\lambda = 2$
Distance separating the position of minimum intensity

= $\frac{\lambda }{2} = \frac{2}{2} = 1\,m$
 

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