In a resonance tube, the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m / s, the frequency of tunning fork is

Question

In a resonance tube, the first resonance with a tuning fork occurs at 16 cm and second at 49 cm. If the velocity of sound is 330 m / s, the frequency of tunning fork is (A) 500 (B) 300 (C) 330 (D) 165

Tardigrade Admin · Accepted Answer

For closed pipe l1=(v/4 n), l2=(3 v/4 n) ⇒ n=(v/2(l2-l1)) =(330/2 ×(0.49-0.16))=500 Hz

Q. In a resonance tube, the first resonance with a tuning fork occurs at $16 c m$ and second at $49 c m$ . If the velocity of sound is $330 m / s$ , the frequency of tunning fork is

500

300

330

165

For closed pipe $l_{1} = \frac{v}{4 n},$
$l_{2} = \frac{3 v}{4 n}$
$\Rightarrow n = \frac{v}{2 ( l _{2} - l _{1} )}$
$= \frac{330}{2 \times ( 0.49 - 0.16 )} = 500 Hz$

Q. In a resonance tube, the first resonance with a tuning fork occurs at 16cm and second at 49cm. If the velocity of sound is 330m/s, the frequency of tunning fork is

500

300

330

165

For closed pipe l1​=4nv​, l2​=4n3v​ ⇒n=2(l2​−l1​)v​ =2×(0.49−0.16)330​=500Hz

Q. In a resonance tube, the first resonance with a tuning fork occurs at $16 c m$ and second at $49 c m$ . If the velocity of sound is $330 m / s$ , the frequency of tunning fork is

For closed pipe $l_{1} = \frac{v}{4 n},$
$l_{2} = \frac{3 v}{4 n}$
$\Rightarrow n = \frac{v}{2 ( l _{2} - l _{1} )}$
$= \frac{330}{2 \times ( 0.49 - 0.16 )} = 500 Hz$