Question

A long glass tube is held vertically, dipping into water, while a tuning fork of frequency $512Hz$ is repeatedly struck and held over the open sent. Strong resonance is obtained, when the length of the tube above the surface of water is $50cm$ and again $84cm$, but not at any intermediate point. Find the speed of sound in air and the next length of the air column for resonance.

Answer 1

Hint: Use the formula for the velocity of sound in air. Substitute the given water in the formula to get the answer.
$v = 2f({l_2} - {l_1})$

Complete step by step answer:
 The glass tube, had vertically, dipping into water will act as a resonance tube.
Velocity of sound in a resonance tube is given by.
$v = 2f({l_2} - {l_1})$
Where,
$v$ is the velocity of sound.
$f$ is the frequency of the tuning fork.
${l_1}$ and ${l_2}$ are the length of tube above the surface of water,
It is given that
$f = 512Hz$
${l_1} = 50cm$
$ = 50cm \times {10^{ - 2}}m(\because 1m = 100cm)$
$ \Rightarrow {l_1} = 0.50cm$.
Simplify, ${l_2} = 84cm = 0.84m$
$\therefore v = 2 + ({l_1} - {l_2})$
By putting the values, we get
$v = 2 \times 512(0.84 - 0.50)$
By simplifying, we get
$v = 1024 \times 0.34$
$ \Rightarrow v = 348.16m/s$
Let the next length for the resonance ${L_e}$${l_{3.}}$
Then, $v = 2f({l_3} - {l_2}).$
By re- arranging, we get
${l_3} - {l_1} = \dfrac{v}{{2f}}$
By substituting the values, we get
${l_3} - 0.84 = \dfrac{{348.16}}{{2 \times 512}}$
By simplifying it, we get
${l_3} = \dfrac{{348.16}}{{1024}} + 0.84$
$ = 0.34 + 0.84$
$ \Rightarrow {l_3} = 1.18m$
$ \Rightarrow {l_3} = 118cm$.
Therefore, the speed of sound in air will be $348.16m/c$
And, the next length of the air column for resonance will be 118cm.

Note:
This is a simple question of substituting values in a formula. You need to focus on substituting the correct values in the correct unit.
Users can log tables to do complex calculations. Learning to use a log table is important.