Question

The mass per unit length of a uniform wire is \[0.135\,gc{m^{ - 1}}\]. A transverse wave of the form \[y = - 0.21\sin \left( {x + 30t} \right)\] is produced in it, where x is in metres and t is in second. Then, the expected value of tension in the wire is \[x \times {10^{ - 2}}N\]. Find the value of x. (Round-off to the nearest integer)

Answer 1

Hint: Tension is defined as a force that acts opposite to the pulling force that is applied on a wire, string, or any one-dimensional material. We might have noticed many objects that are hanged using a rope or cables.

Formula Used:
The formula to find velocity is,
\[v = \sqrt {\dfrac{T}{\mu }} \]
Where, T is tension in the wire and \[\mu \] is mass per unit length.

Complete step by step solution:
Consider a uniform wire having the mass per unit length as \[0.135\,gc{m^{ - 1}}\]and a transverse wave is produced in it. Due to this tension of \[x \times {10^{ - 2}}N\] created in the wire, we need to find the value of x.The formula to find velocity is,
\[v = \sqrt {\dfrac{T}{\mu }} \] ………. (1)
In order to find velocity, we have another formula,
\[v = \dfrac{\omega }{k}\] ………. (2)
Here, the general formula of \[y = - 0.21\sin \left( {x + 30t} \right)\] is,
\[y = A\sin \left( {kx + \omega t} \right)\]
Compare these two equations we get,
\[k = 1\] and \[\omega = 30\]

Now, substitute the value in equation (2) we get,
\[v = \dfrac{{30}}{1}\]
\[\Rightarrow v = 30\,m{s^{ - 1}}\]
Now from equation (2) we can rearrange the equation for tension in the wire as,
\[T = {v^2}\mu \]
Substitute the value of \[v = 30\,m{s^{ - 1}}\] and \[\mu = 0.135\,gc{m^{ - 1}}\]in above equation we get,
\[T = {\left( {30} \right)^2} \times 0.135 \times {10^{ - 1}}\]
\[\Rightarrow T = 900 \times 0.135 \times {10^{ - 1}}\]
\[\Rightarrow T = 12.15\,N\]….. (3)

They have given that the expected value of tension in the wire is \[x \times {10^{ - 2}}N\]
That is, \[T = x \times {10^{ - 2}}N\]……. (4)
Now, equating equations (3) and (4) we get,
\[12.15N = x \times {10^{ - 2}}N\]
\[\Rightarrow x = \dfrac{{12.15}}{{{{10}^{ - 2}}}}\]
\[ \therefore x = 1215\]

Therefore, the value of x is 1215.

Note: Here in this question it is important to remember the units which are about to convert properly. The basic conversion of cm to m and m to cm should be taken care of.