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Question: A wire having a linear density of 0.05g/cm is stretched between two rigid supports with a tension of...

A wire having a linear density of 0.05g/cm is stretched between two rigid supports with a tension of 450N. It is observed that the wire resonated at a frequency of 420 Hz. The next higher frequency at which the same wire resonated is 490Hz. The length of the wire 2142×10−xm2142 \times {10^{ - x}}m. Find x.

Explanation

Solution

Here we are to find the length of the string and we have been given the frequency, tension and linear density. There would be two formulas involved, one where there would be relation between the frequency, velocity and length, second where the relation would be between the tension, velocity and density. Put the second relation in the first and solve for x.

Complete step by step answer:
Here the length of the wire is given as 2142×10−xm2142 \times {10^{ - x}}m. We have to equate the given value to the length of wire to solve for x.
The difference in the frequency is given as:
fo=f2−f1{f_o} = {f_2} - {f_1} ;
Put in the given value of frequencies in the above relation and solve;
⇒fo=490−420\Rightarrow {f_o} = 490 - 420;
⇒fo=70Hz\Rightarrow {f_o} = 70Hz;
The relation between the fundamental frequency fo{f_o}with the velocity v is given as:
fo=v2l{f_o} = \dfrac{v}{{2l}} ; …(Where l = length of the wire)
The relation between velocity and tension in the string is given by:
v=TM/lv = \sqrt {\dfrac{T}{{M/l}}} ;
Here Mass of string / length of string is equal to the density of the string μ\mu .
⇒v=Tμ\Rightarrow v = \sqrt {\dfrac{T}{\mu }};
Put the above relation in the formula fo=v2l{f_o} = \dfrac{v}{{2l}};
fo=Tμ2l{f_o} = \dfrac{{\sqrt {\dfrac{T}{\mu }} }}{{2l}};
Write the above equation in terms of length:
l=Tμ2fol = \dfrac{{\sqrt {\dfrac{T}{\mu }} }}{{2{f_o}}};
Convert the densityμ\mu into kg/m:
μ=0.05×10−310−2kg/m\mu = \dfrac{{0.05 \times {{10}^{ - 3}}}}{{{{10}^{ - 2}}}}kg/m ;
⇒μ=0.05×10−1kg/m\Rightarrow \mu = 0.05 \times {10^{ - 1}}kg/m;
Put the above value in the given below equation:
l=Tμ2fol = \dfrac{{\sqrt {\dfrac{T}{\mu }} }}{{2{f_o}}};
⇒l=4500.05×10−12×70\Rightarrow l = \dfrac{{\sqrt {\dfrac{{450}}{{0.05 \times {{10}^{ - 1}}}}} }}{{2 \times 70}};
Put the given value of l and solve:
2142×10−xm=4500.05×10−11402142 \times {10^{ - x}}m = \dfrac{{\sqrt {\dfrac{{450}}{{0.05 \times {{10}^{ - 1}}}}} }}{{140}};
⇒(2142×10−x)2=4500.05×10−11402\Rightarrow {\left( {2142 \times {{10}^{ - x}}} \right)^2} = \dfrac{{\dfrac{{450}}{{0.05 \times {{10}^{ - 1}}}}}}{{{{140}^2}}}
Simplify the above equation
⇒(2142×10−x)2=45098\Rightarrow {\left( {2142 \times {{10}^{ - x}}} \right)^2} = \dfrac{{450}}{{98}};
⇒(2142×10−x)2=22549\Rightarrow {\left( {2142 \times {{10}^{ - x}}} \right)^2} = \dfrac{{225}}{{49}};
Take the square root on both sides:
⇒2142×10−x=22549\Rightarrow 2142 \times {10^{ - x}} = \sqrt {\dfrac{{225}}{{49}}};
⇒2142×10−x=157\Rightarrow 2142 \times {10^{ - x}} = \dfrac{{15}}{7};
Keep the “x” term on the LHS and take the rest at RHS:
⇒10−x=152142×7\Rightarrow {10^{ - x}} = \dfrac{{15}}{{2142 \times 7}};
⇒10−x=10−3\Rightarrow {10^{ - x}} = {10^{ - 3}};
Now, equate the powers:
⇒−x=−3\Rightarrow - x = - 3;
⇒x=3\Rightarrow x = 3;

Therefore, the value of x is 3.

Note:
It is a length process so go step by step, first find the difference in the frequencies then write the formula for frequency in relation with velocity and length. Here the velocity would be equal to the root of tension upon density. Put the relation of velocity with the tension in the formula for the frequency and find the value of x by equating the given length to the length going to be found.