Question

The speed of a transverse wave, going on a wire having a length of 50 cm and mass of 5.0 g is 80 m/s. The area of the cross-section of the wire is \[1m{m^2}\]. Young's modulus is \[1.6 \times {10^{11}}\]. Find the extension of the wire over its natural length.

Answer 1

Hint: First of all, determine the linear mass density of the wire, and then using the relation between the speed of the wave and tension, we will determine the tension developed in the wire. After that apply the relation between the young’s modulus and the strain. Hence, we will get a suitable answer.

Formula used:
\[E = \dfrac{\sigma }{\varepsilon }\]

Complete step by step solution:
In the above problem, we have given the mass of the wire, length of the wire, and cross-section of the wire.
Therefore,
Mass of the wire (m) = 5 Gram, length of the wire(L) = 50cm, cross section of the wire (A) = \[1m{m^2}\] and
Young’ modulus = \[1.6 \times {10^{11}}\] Speed of the wave = 80m/s
Now we have to determine the extension of the wire over its natural length.
Therefore,
We know that
\[\begin{array}{*{20}{c}}{ \Rightarrow E}& = &{\dfrac{\sigma }{\varepsilon }}\end{array}\]
Where
E = young’s modulus
\[\sigma \]= stress induced due to tensile force
\[\varepsilon \]= strain
We know that
\[\begin{array}{*{20}{c}}{ \Rightarrow \sigma }& = &{\dfrac{T}{A}}\end{array}\] And \[\begin{array}{*{20}{c}}{ \Rightarrow \varepsilon }& = &{\dfrac{{\Delta L}}{L}}\end{array}\]
Where
T = tensile force
\[\Delta L\]= extension in the wire
Therefore,
\[\begin{array}{*{20}{c}}{ \Rightarrow E}& = &{\dfrac{{\dfrac{T}{A}}}{{\dfrac{{\Delta L}}{L}}}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow E}& = &{\dfrac{{TL}}{{A\Delta L}}}\end{array}\] ……. (1).
Now, we know that the linear mass density is,
\[\begin{array}{*{20}{c}}{ \Rightarrow \mu }& = &{\dfrac{m}{L}}\end{array}\]
Therefore, put the value
\[\begin{array}{*{20}{c}}{ \Rightarrow \mu }& = &{\dfrac{{5 \times {{10}^{ - 3}}}}{{5 \times {{10}^{ - 2}}}}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow \mu }& = &{0.1{\raise0.5ex\hbox{$\scriptstyle {kg}$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle m$}}}\end{array}\]
So, the tensile force is given by
\[\begin{array}{*{20}{c}}{ \Rightarrow v}& = &{\sqrt {\dfrac{T}{\mu }} }\end{array}\]
Put the value,
\[\begin{array}{*{20}{c}}{ \Rightarrow 80}& = &{\sqrt {\dfrac{T}{{0.1}}} }\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow T}& = &{64N}\end{array}\]
Now put all the values in the equation (1). Therefore, we will get
\[\begin{array}{*{20}{c}}{ \Rightarrow E}& = &{\dfrac{{TL}}{{A\Delta L}}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow 1.6 \times {{10}^{11}}}& = &{\dfrac{{64 \times 0.5}}{{1 \times {{10}^{ - 6}} \times \Delta L}}}\end{array}\]
\[\begin{array}{*{20}{c}}{ \Rightarrow \Delta L}& = &{0.02mm}\end{array}\]

So, the extension in the wire is 0.02mm.

Note: It is important to note that the units of all the parameters must be the same otherwise there will be an error in the solution. The common errors occur in reading, zero error, etc. These problems while solving tend to make many mistakes because it includes a number of formulas to remember.