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प्रश्न
In a resonance column experiment, a tuning fork of frequency 400 Hz is used. The first resonance is observed when the air column has a length of 20.0 cm and the second resonance is observed when the air column has a length of 62.0 cm. (a) Find the speed of sound in air. (b) How much distance above the open end does the pressure node form?
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उत्तर
Given:
Length of air column at first resonance L1 = 20 cm = 0.2 m
Length of air column at second resonance L2 = 62 cm = 0.62 m
Frequency of tuning fork f = 400 Hz
(a) We know that :
\[\lambda = 2\left( L_2 - L_1 \right)\]
\[ \Rightarrow \lambda = 2\left( 62 - 20 \right) = 84 \text { cm } = 0 . 84 \text{ m }\]
v = λf,
where v is the speed of the sound in air.
So,
\[v = 0 . 84 \times 400 = 336 \text{ m/s }\]
Therefore, the speed of the sound in air is 336 m/s.
(b) Distance of open node is d :
\[L_1 + d = \frac{\lambda}{4}\]
\[ \Rightarrow d = \frac{\lambda}{4} - L_1 = 21 - 20 = 1 \text { cm }\]
Therefore, the required distance is 1 cm.
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