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प्रश्न
A tuning fork of unknown frequency makes 5 beats per second with another tuning fork which can cause a closed organ pipe of length 40 cm to vibrate in its fundamental mode. The beat frequency decreases when the first tuning fork is slightly loaded with wax. Find its original frequency. The speed of sound in air is 320 m s−1.
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उत्तर
Given:
Length of the closed organ pipe L = 40 cm = 40 × 10−2 m
Velocity of sound in air v = 320 ms−1
Frequency of the fundamental note of a closed organ pipe \[\left( n \right)\] is given by : \[n = \frac{v}{4L}\]
⇒ \[n = \frac{v}{4L} = \frac{320}{4 \times 40 \times {10}^{- 2}} = 200 \text{ Hz }\]
As the tuning fork produces 5 beats with the closed pipe, its frequency must be 195 Hz or 205 Hz.
The frequency of the tuning fork decreases as and when it is loaded. Therefore, the frequency of the tuning fork should be 205 Hz.
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