Question

Length of an elastic string is $x$ metre when longitudinal tension is $3 \,N$ and $y$ metre when it is $5 \,N$. If length of string is found to be $2.5 y-1.5 x$, calculate tension in the string in $N$.

Answer 1

Let $L$ be original length of string and $k$ be force constant of string.
New final length $=$ initial length $+$ elongation
$L^{\prime}=L+\frac{F}{K}$
When tension of $3 N$ is applied,
$x = L +\frac{3}{ k } ....$ (i)
When tension of $5 N$ is applied,
$y=L+\frac{5}{k} ....$(ii)
Subtracting equation (ii) from (i),
$ x-y=\frac{3}{k}-\frac{5}{k} $
$\therefore k=\frac{-2}{x-y}=\frac{2}{y-x}$
Substituting value of $k$ in equation (i),
$ x=L+\frac{3}{2 / y-x}=L+\frac{3(y-x)}{2}$
$\therefore L=x-\frac{3(y-x)}{2}$
$=\frac{2 x-3 y+3 x}{2}=\frac{5 x-3 y}{2}$
Let $T$ be tension when length is $2.5 y -1.5 x$
$ \therefore 2.5 y -1.5 x = L +\frac{ T }{ k } $
$ 2.5 y -1.5 x =\frac{5 x -3 y }{2}+\frac{ T }{(2 / y - x )} $
$ 2.5 y -1.5 x =\frac{5 x -3 y }{2}+\frac{( y - x ) T }{2}$
$\therefore 5 y -3 x =5 x -3 y +( y - x ) T$
$ \therefore 8 y -8 x =( y - x ) T$
$\therefore 8( y - x )=( y - x ) T $
$\Rightarrow T =8 \,N $