Question

A uniform spring has an unstretched length $L$ and a force constant $k$ . The spring is cut into two parts of unstretched lengths $l_{1}$ and $l_{2}$ such that $l_{2}=\eta l_{2}$ , where $\eta$ is an integer. The corresponding force constants are $k_{1}$ and $k_{2}$ , then

• A. force has to be kept same to find $k_{1}$ and $k_{2}$
• B. $k_{1}=\frac{k \left(\eta + 1\right)}{\eta}$ and $k_{2}=k\left(\eta - 1\right)$
• C. $k_{1}=\frac{k \left(\eta - 1\right)}{\eta}$ and $k_{2}=k\left(\eta + 1\right)$
• D. $k_{1}=\frac{k \left(\eta + 1\right)}{\eta}$ and $k_{2}=k\left(\eta + 1\right)$

Answer 1

Answer: (D)

In case of cutting of a spring, $l\times k=constant$
$k_{1}l_{1}=k_{2}l_{2}$
$k_{1}\left(\eta l_{2}\right)=k_{2}l_{2} \, \Rightarrow \eta k_{1}=k_{2}$ ...............(1)
Now both the springs when connected in series should give the same effect as that of the original spring. Hence, by applying the concept of springs connected in series, we obtain,
$\frac{1}{k}=\frac{1}{k_{1}}+\frac{1}{k_{2}}$ ...............(2)
$\frac{1}{k}=\frac{1}{k_{1}}+\frac{1}{\eta k_{1}} \, \, \, \Rightarrow \frac{k_{1}}{k}=\frac{\eta + 1}{\eta}$ (using (1))
$k_{1}=\left(\frac{\eta + 1}{\eta}\right)k$ ............(3)
Using (3) again in (2), we obtain,
$k_{2}=\left(\eta + 1\right)k$