When a vibrating tuning fork of frequency 512 Hz is held above the mouth of a resonance tube of adjustable…
When a vibrating tuning fork of frequency 512 Hz is held above the mouth of a resonance tube of adjustable length, the first two successive positions of resonance occur when the length of the air columns are 15.4 cm and 48.6 cm, respectively. Then, the velocity of sound is
(a) $512 (48.6 - 15.4)\text{ cm s}^{-1}$
(b) $1024 (48.6 - 15.4)\text{ cm s}^{-1}$
(c) $256 (48.6 - 15.4)\text{ cm s}^{-1}$
(d) $2 \times 512 (48.6 + 15.4)\text{ cm s}^{-1}$
Solution
Difference in lengths will be equal to one loop or $\frac{\lambda}{2}$.
$\cdot\!\cdot\!\cdot\, \lambda = 2(48.6 - 15.4)\text{ cm}$
Now, $v = f \lambda = 512 \times 2(48.6 - 15.4)$
$= 1024(48.6 - 15.4)\text{ cm s}^{-1}$