Question

A transverse wave travels on a taut steel wire with a velocity of V when tension in it is \[2.06 \times {10^4}N\]. When the tension is changed to T, the velocity changes to \[\dfrac{V}{2}\]. The value of T is close to
A. \[30.5 \times {10^4}\]
B. \[2.50 \times {10^4}\]
C. \[10.2 \times {10^2}\]
D. \[5.15 \times {10^3}\]

Answer 1

Hint: When the direction of a wave and its motion of vibration are orthogonal or at an angle of 90 degrees, then those waves are known as transverse waves. Tension is termed as the force in a wire when it is stretched or pulled from two ends and it is generally taken in the middle of both the ends from where the force is applied.

Formula Used:
\[V = \sqrt {\dfrac{T}{\mu }} \]
where V is the velocity in a medium, T is the tension and \[\mu \] is linear density.

Complete step by step solution:
We have to find the value of tension for which the velocity V becomes half of the original velocity. We can find it by finding the relation between the tension of the wire and the velocity of the wave. Let the original velocity be \[{V_1} = V\] and the changed velocity be \[{V_2} = \dfrac{V}{2}\], and let the original tension be \[{T_1} = 2.06 \times {10^4}N\] and the changed tension be \[{T_2} = T\]. We know that mass and length of the wire are constant so its linear density is also constant.

We can find the relation between velocity and tension as,
\[V = \sqrt {\dfrac{T}{\mu }} \\
\Rightarrow \dfrac{{{V_1}}}{{{V_2}}} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}} \\
\Rightarrow \dfrac{V}{{\left( {\dfrac{V}{2}} \right)}} = \sqrt {\dfrac{{2.06 \times {{10}^4}}}{T}} \\
\Rightarrow 2 = \sqrt {\dfrac{{2.06 \times {{10}^4}}}{T}} \]
Squaring both sides, we get,
\[4 = \dfrac{{2.06 \times {{10}^4}}}{T} \\
\Rightarrow T = \dfrac{{2.06 \times {{10}^4}}}{4} \\
\Rightarrow T = 0.515 \times {10^4} \\
\Rightarrow T = 5.15 \times {10^3}N \]

So, option D i.e, \[5.15 \times {10^3}\] is the required solution.

Note: Tension is created in the wire when it is stretched, so even if the wire is stretched it does not mean that its length is also changed. The original length of the wire will remain constant and so the linear density also remains constant. As we don’t know the value of linear density, we can remove it by dividing the velocity of the wave in wire with the original tension, with the velocity of the wave in wire with the changed tension.