Question

The fundamental frequency of sonometer wire carrying a block of mass $1\,kg$ and density $1.8$ is $260.$ when the block is completely immersed in liquid of density $1.2$ then what will be its new frequency?
A. $300\,Hz$
B. $150\,Hz$
C. $450\,Hz$
D. None of these

Answer 1

Hint:In order to solve this question we need to understand sound propagation, buoyant force and sonometer. Sound wave is a disturbance in a medium which propagates as consecutive compression and rarefaction in a medium; it requires a material medium to transfer or propagate from one place to another. Sonometer is an instrument that is used to diagnose tension, frequency and density of vibrations, generally used in medicine to test both bones and hearing density. Buoyant force is that reaction force of water when it is pressed by some object.

Complete step by step answer:
Let the length of the sonometer tube be, $L$ and also let the wavelength and frequency denoted as $\lambda \& f$ respectively. Also let the linear density of string through which mass is attached be $\mu $. Since we know, sonometer tube is both end closed organ pipe so there are two nodes at boundary, hence length of sonometer tube is defined as,
$L = \dfrac{{n\lambda }}{2}$
Since waves vibrate in fundamental frequency, $n = 1$.
So, $L = \dfrac{\lambda }{2}$
Or wavelength is, $\lambda = 2L \to (i)$

Let the wave speed be, $v$. So the relation between frequency, wavelength and wave speed is, $v = f\lambda $. Let tension in string be, $T$. Since wave speed is defined in terms of tension and linear density as, $v = \sqrt {\dfrac{T}{\mu }} $.
So the frequency is defined as, $f = \dfrac{v}{\lambda }$
Putting values we get, $f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} $
So for first case when mass is not dipped inside liquid then, $T = mg$
Here, “m” is mass of the body and “g” is acceleration due to gravity.
Also let the density of mass be, $\rho = 1.8$ (given) and Volume of mass be, “V”.
So mass defined in terms of density and volume as, $m = \rho Vg$
Putting value of “m” to get tension as, $T = V\rho g$
So frequency is, $f = \dfrac{1}{{2L}}\sqrt {\dfrac{{V\rho g}}{\mu }} $
Also value of frequency given when not dipped in liquid is, $f = 260Hz$
So, putting value of density and frequency as,
$260 = \dfrac{1}{{2L}}\sqrt {\dfrac{{1.8Vg}}{\mu }} \to (ii)$

Now when the mass dipped inside liquid then buoyant force acts in upward direction, so now force equation become,
$T' + {F_b} = mg$
$\Rightarrow T' = mg - {F_B}$
Since buoyant force is given as, ${F_b} = V{\rho _l}g$
Here, ${\rho _l}$ is the density in liquid given as, ${\rho _l} = 1.2$
Putting value of mass and buoyant force we get tension as,
$T' = V\rho g - V{\rho _l}g$
$\Rightarrow T' = 1.8Vg - 1.2Vg$
$\Rightarrow T' = 0.6Vg$

Also let the new frequency by which the wave vibrates is, $f'$. So new frequency in terms of tension as,
$f' = \dfrac{1}{{2L}}\sqrt {\dfrac{{T'}}{\mu }} $
$\Rightarrow f' = \dfrac{1}{{2L}}\sqrt {\dfrac{{0.6Vg}}{\mu }} \to (iii)$
Now dividing equation (iii) by (ii) we get,
$\dfrac{{f'}}{{260}} = \dfrac{{\sqrt {\dfrac{{1.8Vg}}{\mu }} }}{{\sqrt {\dfrac{{0.6Vg}}{\mu }} }}$
$\Rightarrow \dfrac{{f'}}{{260}} = \sqrt {\dfrac{{1.8Vg}}{{0.6Vg}}} $
$\Rightarrow \dfrac{{f'}}{{260}} = \sqrt {\dfrac{1}{3}} $
$\Rightarrow f' = \dfrac{{260}}{{\sqrt 3 }}$
$ \therefore f' = 150.12Hz$

The correct option is B.

Note:It should be remembered that, block is completely immersed so in expression of buoyant force complete volume is written otherwise buoyant force acts only on those masses of liquid which are displaced by the object. Also linear density of rope is defined for distribution of mass on rope, linear heat means rope is one dimensional.