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प्रश्न
A source of sound S and detector D are placed at some distance from one another. a big cardboard is placed near hte detector and perpendicular to the line SD as shown in figure. It is gradually moved away and it is found that the intensity changes from a maximum to a minimum as the board is moved through a distance of 20 cm. Find the frequency of the sound emitted. Velocity of sound in air is 336 m s−1.
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उत्तर
Given:
Velocity of sound in air v = 336 ms−1
Distance between maximum and minimum intensity :
\[\frac{\lambda}{4}\]= 20 cm
Frequency of sound f = ?
We have :
\[\frac{\lambda}{4} = 20\]
\[ \Rightarrow \lambda = 20 \times 4 = 80 \text { cm } = 80 \times {10}^{- 2} \text { m }\]
As we know ,
\[v = f\lambda\]
\[\therefore\] \[f = \frac{v}{\lambda}\]
\[\Rightarrow f = \frac{336}{80 \times {10}^{- 2}} = 420 \text { Hz }\]
Therefore, the frequency of the sound emitted from the source is 420 Hz.
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