First slide

Stationary waves

Question

An open tube is in resonance with string (frequency of vibration of tube is n₀). If tube is dipped in water so that 75% of length of tube is inside water, then the ratio of the frequency of tube to string now will be

Moderate

A

1
B

2
C

$\frac{2}{3}$
D

$\frac{3}{2}$

Solution

For open tube, $n_{0} = \frac{v}{2 l}$
For closed tube length available for resonance is
$l' = l \times \frac{25}{100} = \frac{l}{4} ∴$ Fundamental frequency of water filled tube $n = \frac{v}{4 l'} = \frac{v}{4 \times (l / 4)} = \frac{v}{l} = 2 n_{0} \Rightarrow \frac{n}{n_{0}} = 2$

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