Question

A uniform spring has an unstretched length $l$ and a force constant $k$. The spring is cut into two parts of unstretched length $l_1$ and $l_2$ such that $l_1=\eta l_2$, where $\eta$ is an integer. The corresponding force constants $k_1$ and $k_2$ are

• A. $k\eta$ and $k(\eta +1)$
• B. $\frac{k(\eta +1)}{\eta }$ and $k(\eta -1)$
• C. $\frac{k(\eta -1)}{\eta }$ and $k(\eta +1)$
• D. $\frac{k(\eta +1)}{\eta }$ and $k(\eta +1)$

Answer 1

The correct option is D $\frac{k(\eta +1)}{\eta }$ and $k(\eta +1)$

If a mass $m$ is hung from the lower end of an unstretched length $l$ of the spring and $x_0$ is the extensions produces in the string, then

$k{x}_0+mg=0$

or

$k=\frac{mg}{x_0}$

Now,

$x_0\propto l$

or

$x_0=cl$

$k=\frac{mg}{cl}$

As the spring is cut into two pieces of lengths $l_1$ and $l_2$, so $l_1+l_2=l$

As,

$l_1=\eta l_2$

$l_2=\frac{l}{\eta +1}$ and $\frac{\eta l}{\eta +l}$

Force constants $k_1$ and $k_2$ for pieces of length $l_1$ and $l_2$ are:

$k_1=\frac{mg}{c{l}_1}=\frac{mg}{c}\left(\frac{\eta +1}{\eta l}\right)=\frac{mg}{cl}\left(\frac{\eta +1}{\eta l}\right)=k\frac{(\eta +1)}{\eta }$

and

$k_2=\frac{mg}{c{l}_2}=\frac{mg}{c}\frac{(\eta +1)}{l}=k(\eta +1)$