Question

An object of specific gravity $p$ is hung from a thin steel wire. The fundamental frequency for transverse standing waves in the wire is$300Hz$. The object is immersed in water so that one half of its volume is submerged. The new fundamental frequency is $Hz$ is
(A) $300{(\dfrac{{2p - 1}}{{2p}})^{1/2}}$
(B) $300{(\dfrac{{2p}}{{2p - 1}})^{1/2}}$
(C) $300(\dfrac{{2p}}{{2p - 1}})$
(D) $300(\dfrac{{2p - 1}}{{2p}})$

Answer 1

Hint:-
- Fundamental frequency is given by the equation$f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{M}}$.
- The immersed object will get an upthrust from the water.

Complete step by step solution:-
The fundamental frequency is $f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{M}}$
Where $L$ is the length of the string.
$M$ is the linear mass density of the string.
$T$ is the tension of the string.
The fundamental frequency for transverse standing waves in the wire at air is given is $300Hz$.
Consider the tension $T$ on the string in air. If an object is immersed in a liquid the tension is changed to$T'$.
Consider $V$ is the volume of liquid, $p$is the specific gravity, $g$ is the acceleration due to gravity.
$T' = T -$(Upthrust force).
Here only the half of the volume is displaced, so that
Mass of the volume displace is half of the product of volume and density
$= \dfrac{1}{2}V\rho$
$\rho$is the density of liquid.
The upthrust force $= \dfrac{1}{2}V\rho g$
$T' = T - \dfrac{1}{2}V\rho g$
Tension in the air $T = Vpg$
Density of water $\rho = 1g/cc$
$T' = Vpg - \dfrac{1}{2}Vg$
$T' = Vg(p - \dfrac{1}{2})$
New fundamental frequency, after the object is immersed in water,
$f' = \dfrac{1}{{2L}}\sqrt {\dfrac{{T'}}{M}}$
Substitute new tension value in this equation we get
$f' = \dfrac{1}{{2L}}\sqrt {\dfrac{{Vg(p - \dfrac{1}{2})}}{M}}$
Compare this with the fundamental frequency before the object immersed in water,

$$f'/300 = \sqrt {\dfrac{{p - \dfrac{1}{2}}}{P}} \dfrac{{\sqrt {(p - \dfrac{1}{2})} }}{{\sqrt p }} \\\ \\\$$

Simplify the equation,
$f'/f = \dfrac{{\sqrt {Vg(p - \dfrac{1}{2})} }}{{\sqrt {Vpg} }}$
The fundamental frequency ($f$) is already given, $f = 300Hz$
$f'/300 = \dfrac{{\sqrt {Vg(p - \dfrac{1}{2})} }}{{\sqrt {Vpg} }}$
We cancel the terms inside the root also,
$f'/300 = \dfrac{{\sqrt {(p - \dfrac{1}{2})} }}{{\sqrt p }}$
$f'/300 = \sqrt {\dfrac{{p - \dfrac{1}{2}}}{P}}$
Now rearrange the equation, like the given options,

$f'/300 = {(\dfrac{{p - \dfrac{1}{2}}}{P})^{1/2}}$
$f' = 300{(\dfrac{{2p - 1}}{{2P}})^{1/2}}$

So the answer is (A) $f' = 300{(\dfrac{{2p - 1}}{{2P}})^{1/2}}$

Note:-
- The value of acceleration due to gravity is $g = 9.8m/s$.
- The unit of frequency is Hertz ($Hz$).
- The sound in the water is faster than sound in air.
- The immersed object weight is equivalent to the volume displaced times the density and acceleration due to gravity.