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प्रश्न
In Quincke's experiment the sound detected is changed from a maximum to a minimum when the sliding tube is moved through a distance of 2.50 cm. Find the frequency of sound if the speed of sound in air is 340 m s−1.
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उत्तर
Given:
Speed of sound in air v = 340 ms−1
Distance moved by sliding tube = 2.50 cm
Frequency of sound f = ?
\[\text { Distance between maximum and minimum: } \] \[ \frac{\lambda}{4} = 2 . 50 \text { cm }\]
\[ \Rightarrow \lambda = 2 . 50 \times 4 = 10 \text { cm } = {10}^{- 1} \text { m }\]
As we know,
v = f\[\lambda\].
\[\therefore f = \frac{v}{\lambda}\]
\[ \Rightarrow f = \frac{340}{{10}^{- 1}}=3400\text { Hz }=3.4\text{ kHz }\]
Therefore, the frequency of the sound is 3.4 kHz.
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