Question

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

Answer 1

Answer: 8

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 $3N$ is applied,
$x=L+\frac{3}{k}....$ (i)
When tension of $5N$ 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),
$x=L+\frac{3}{2/y-x}=L+\frac{3(y-x)}{2}$
$\therefore L=x-\frac{3(y-x)}{2}$
$=\frac{2x-3y+3x}{2}=\frac{5x-3y}{2}$
Let $T$ be tension when length is $2.5y-1.5x$
$\therefore 2.5y-1.5x=L+\frac{T}{k}$
$2.5y-1.5x=\frac{5x-3y}{2}+\frac{T}{(2/y-x)}$
$2.5y-1.5x=\frac{5x-3y}{2}+\frac{(y-x)T}{2}$
$\therefore 5y-3x=5x-3y+(y-x)T$
$\therefore 8y-8x=(y-x)T$
$\therefore 8(y-x)=(y-x)T$
$\Rightarrow T=8N$