A wire of length one meter under a certain initial tension emits a sound of fundamental frequency \[256\,{\text{Hz}}\]. When the tension is increased by \[1\,{\text{kgwt}}\], the frequency of the frequency of the fundamental node increases to \[320\,{\text{Hz}}\]. The initial tension is
A. \[3/4\,{\text{kgwt}}\]
B. \[4/3\,{\text{kgwt}}\]
C. \[16/9\,{\text{kgwt}}\]
D. \[20/9\,{\text{kgwt}}\]
Answer
642.6k+ views
Hint:Use the formula of fundamental frequency in a stretched string. This formula gives the relation between the fundamental frequency of vibration of a stretched string, tension in the stretched string, length of the stretched string and linear density of the stretched string. Deduce the relation between the fundamental frequency and tension in the string to solve the question.
Formula used:
The fundamental frequency \[f\] of a stretched string is given by
\[f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} \] …… (1)
Here, \[L\] is the length of the stretched string, \[T\] is the tension in the string and \[\mu \] is the linear density of the string.
Complete step by step answer:
We have given that the initial fundamental frequency \[f\] of the wire having length \[1\,{\text{m}}\] is \[256\,{\text{Hz}}\] and the increased fundamental frequency \[f'\] when the tension in the wire is increased by \[1\,{\text{kgwt}}\] is \[320\,{\text{Hz}}\].
\[L = 1\,{\text{m}}\]
\[\Rightarrow f = 256\,{\text{Hz}}\]
\[\Rightarrow f' = 320\,{\text{Hz}}\]
Let the initial tension in the string is \[T\,{\text{kgwt}}\] then the increased tension will be \[\left( {T + 1} \right)\,{\text{kgwt}}\].
From equation (1). We can conclude that the fundamental frequency of the wire is directly proportional to the square root of the tension in the wire and inversely proportional to the length of the wire and the square root of the linear density of the wire.
In the present case of wire, the length of the wire and the linear density of the wire remains the same. Hence, the fundamental frequency of the wire is directly proportional to the square root of the tension in the wire.
\[f \propto \sqrt T \]
Write the above relation between the fundamental frequency and tension in the wire for the initial and final condition.
\[\dfrac{f}{{f'}} = \sqrt {\dfrac{T}{{T + 1}}} \]
Take square on both sides of the above equation.
\[\dfrac{{{f^2}}}{{f{'^2}}} = \dfrac{T}{{T + 1}}\]
Substitute \[256\,{\text{Hz}}\] for \[f\] and \[320\,{\text{Hz}}\] for \[f'\] in the above equation.
\[\dfrac{{{{\left( {256\,{\text{Hz}}} \right)}^2}}}{{{{\left( {320\,{\text{Hz}}} \right)}^2}}} = \dfrac{T}{{T + 1}}\]
\[ \Rightarrow \dfrac{{T + 1}}{T} = \dfrac{{{{\left( {320\,{\text{Hz}}} \right)}^2}}}{{{{\left( {256\,{\text{Hz}}} \right)}^2}}}\]
\[ \Rightarrow 1 + \dfrac{1}{T} = \dfrac{{{{\left( {320\,{\text{Hz}}} \right)}^2}}}{{{{\left( {256\,{\text{Hz}}} \right)}^2}}}\]
\[ \Rightarrow \dfrac{1}{T} = 1.5625 - 1\]
\[ \Rightarrow \dfrac{1}{T} = 0.5625\]
\[ \therefore T = \dfrac{{16}}{9}\,{\text{kgwt}}\]
Therefore, the initial tension in the wire is \[\dfrac{{16}}{9}\,{\text{kgwt}}\].
Hence, the correct option is C.
Note:The students may think that the unit of the increase in tension in the wire is given kgwt in the question and it is not converted in the SI unit of the tension force which is newton. But the unit kgwt is equivalent to the unit newton of the force. So, it can be used instead of kgwt.
Formula used:
The fundamental frequency \[f\] of a stretched string is given by
\[f = \dfrac{1}{{2L}}\sqrt {\dfrac{T}{\mu }} \] …… (1)
Here, \[L\] is the length of the stretched string, \[T\] is the tension in the string and \[\mu \] is the linear density of the string.
Complete step by step answer:
We have given that the initial fundamental frequency \[f\] of the wire having length \[1\,{\text{m}}\] is \[256\,{\text{Hz}}\] and the increased fundamental frequency \[f'\] when the tension in the wire is increased by \[1\,{\text{kgwt}}\] is \[320\,{\text{Hz}}\].
\[L = 1\,{\text{m}}\]
\[\Rightarrow f = 256\,{\text{Hz}}\]
\[\Rightarrow f' = 320\,{\text{Hz}}\]
Let the initial tension in the string is \[T\,{\text{kgwt}}\] then the increased tension will be \[\left( {T + 1} \right)\,{\text{kgwt}}\].
From equation (1). We can conclude that the fundamental frequency of the wire is directly proportional to the square root of the tension in the wire and inversely proportional to the length of the wire and the square root of the linear density of the wire.
In the present case of wire, the length of the wire and the linear density of the wire remains the same. Hence, the fundamental frequency of the wire is directly proportional to the square root of the tension in the wire.
\[f \propto \sqrt T \]
Write the above relation between the fundamental frequency and tension in the wire for the initial and final condition.
\[\dfrac{f}{{f'}} = \sqrt {\dfrac{T}{{T + 1}}} \]
Take square on both sides of the above equation.
\[\dfrac{{{f^2}}}{{f{'^2}}} = \dfrac{T}{{T + 1}}\]
Substitute \[256\,{\text{Hz}}\] for \[f\] and \[320\,{\text{Hz}}\] for \[f'\] in the above equation.
\[\dfrac{{{{\left( {256\,{\text{Hz}}} \right)}^2}}}{{{{\left( {320\,{\text{Hz}}} \right)}^2}}} = \dfrac{T}{{T + 1}}\]
\[ \Rightarrow \dfrac{{T + 1}}{T} = \dfrac{{{{\left( {320\,{\text{Hz}}} \right)}^2}}}{{{{\left( {256\,{\text{Hz}}} \right)}^2}}}\]
\[ \Rightarrow 1 + \dfrac{1}{T} = \dfrac{{{{\left( {320\,{\text{Hz}}} \right)}^2}}}{{{{\left( {256\,{\text{Hz}}} \right)}^2}}}\]
\[ \Rightarrow \dfrac{1}{T} = 1.5625 - 1\]
\[ \Rightarrow \dfrac{1}{T} = 0.5625\]
\[ \therefore T = \dfrac{{16}}{9}\,{\text{kgwt}}\]
Therefore, the initial tension in the wire is \[\dfrac{{16}}{9}\,{\text{kgwt}}\].
Hence, the correct option is C.
Note:The students may think that the unit of the increase in tension in the wire is given kgwt in the question and it is not converted in the SI unit of the tension force which is newton. But the unit kgwt is equivalent to the unit newton of the force. So, it can be used instead of kgwt.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

