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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) \[300(\dfrac{{2p - 1}}{{2p}})\]

seo-qna
Last updated date: 16th Jun 2024
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Answer
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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.