Question

A tube of certain diameter $'d'$ and length $48\,cm$is open at both ends. Its fundamental frequency of resonance is found to be $320\,Hz$ and the speed of sound in air is $320\,m{s^{ - 1}}$. Estimate the diameter of the tube. If one end of the tube is closed then, calculate the lowest frequency of resonance for the tube.

Answer 1

Hint
In the question, frequency and length is given. By substituting the value in the frequency of resonance formula. We get the value of radius and diameter and using this we get the value of resonance. The expression for finding wavelength is
$\dfrac{\lambda }{2} = l + 1.2\,r$
Where, $\lambda$ be the wavelength, $r$ be the radius of the tube and $l$ be the length of the tube.
The expression for finding frequency of resonance is
$f = \dfrac{v}{\lambda }$
Where, $v$ be the speed of the sound (velocity) , $\lambda$ be the wavelength and $f$ be the frequency of the resonance.

Complete Step by Solution
We know that,
Length $l$= $48\,cm$, speed of the sound $v = 320\,m{s^{ - 1}}$ , frequency of resonance $f = 320\,Hz$
$\dfrac{\lambda }{2} = l + 1.2\,r...........\left( 1 \right)$
$\lambda = \dfrac{v}{f}............\left( 2 \right)$
Substitute the known values in the equation $\left( 2 \right)$,
$\lambda = \dfrac{{320}}{{320}}$
$\lambda = 1\,m$
$\lambda = 100\,cm$
$\dfrac{\lambda }{2} = l + 1.2\,r$, Substitute the known values in the equation $\left( 1 \right)$,
Therefore, $\dfrac{{100}}{2} = 48 + 1.2\,r$
Simplify the above equation we get the value of $r$,
$r = 1.67\,cm$
or $d = 2 \times radius = 2\,r$
So $d = 2 \times 1.67 = 3.33\,cm$
From the question, we know that one end of the tube is closed, then the equation is,
$\dfrac{\lambda }{4} = l + 0.6\,r$
$\dfrac{\lambda }{4} = 48 + 0.6 \times 1.67$
Therefore,
$\lambda = 196\,cm$.
We know that frequency $f = \dfrac{v}{\lambda }$
Substitute the value of $v$and $\lambda$ in the above equation,
$f = \dfrac{{320 \times 100}}{{196}}$
Hence, the frequency of resonance of one side of the tube is $f = 163.3\,Hz$.

Note
By the question, first we have to find the value of frequency of resonance of two sides. Then we have to find the value of frequency of resonance on one side of the tube. Substituting value of velocity and wavelength we can find the frequency of resonance.