Question

When a mass $M$ hangs from a spring of length $L$, it stretches the spring by a distance $x$. Now the spring is cut in two parts of length $\frac{L}{3}$ and $\frac{2L}{3}$, and the two springs thus formed are connected to a straight rod of mass $M$ which is horizontal in the configuration as shown in the figure. Find the stretch in each of the spring.

• A. $x_1=\frac{x}{6},x_2=\frac{x}{2}$
• B. $x_1=\frac{x}{3},x_2=\frac{x}{6}$
• C. $x_1=\frac{x}{2},x_2=\frac{x}{2}$
• D. $x_1=\frac{x}{6},x_2=\frac{x}{3}$

Answer 1

The correct option is D $x_1=\frac{x}{6},x_2=\frac{x}{3}$

As it given that the mass $M$ stretches the original spring by a distance $x$, we have
$kx\text{=}Mg$
or $k=\frac{Mg}{x}$.....(1)
The new spring constants of the two springs can be obtained as,
$k_1=3k$ and $k_2=\frac{3k}{2}$[Since $k_1{L}_1=k_2{L}_2=kL]$
Let us take the stretch in the two springs to be $x_1$ and $x_2$, we have for the equilibrium of the rod.
$k_1{x}_1+k_2{x}_2=Mg$
or $3k{x}_1+\frac{3k}{2}{x}_2=Mg$......(2)
From equation (1) we have
$x_1+\frac{x_2}{2}=\frac{x}{3}$......(3)
As the rod is horizontal and in static equilibrium, so torque acting on the rod about any point on it must be zero.
Thus we have torque on it about end A
$k_2{x}_{2}L=Mg\frac{L}{2}$
or $x_2=\frac{Mg}{2k_2}=\frac{kx}{2\left[\frac{3k}{2}\right]}[\text{Using}Mg=kx\text{and}{k}_2=\frac{3k}{2}]$
$x_2=\frac{x}{3}$......(4)
Using this value of $x_2$ in equation (3), we have $x_1=\frac{x}{6}$
This can also be directly obtained by putting torque equal to zero about point B on the rod:
$k_1{x}_{1}L=Mg\frac{L}{2}$
or $x_1=\frac{Mg}{2k_1}$
or $\frac{Mg}{6k}=\frac{x}{6}$ $[\text{Using}{x}_1=\frac{x}{6},k_1=3k]$