Question

A tube of certain diameter and of length 48 cm is open at both ends. Its fundamental frequency of resonance is found to be 320 Hz. If velocity of sound in air is $320m{s}^{-1}$ the diameter of the tube is

• A. 1.33cm
• B. 2.33cm
• C. 3.33cm
• D. 4.33cm

Answer 1

Answer: (C)

$\frac{\lambda }{2}=48+2(0.3d)$
where .d is diameter of the tube
but $\lambda =\frac{V}{n}=\frac{320}{320}=lm=100cm$
$\therefore \frac{100}{2}=48+(0.6)d\Rightarrow d=\frac{2}{0.6}=3.33cm$