Question

A block of mass $2kg$ is connected to one end of a system of springs as shown in figure. The springs are massless, having spring constants $k_1=100N/m,k_2=200N/m$, and natural lengths of $0.5m$ each. The other end of the system is fixed at the centre of a frictionless horizontal table. Initially, spring mass system is at rest. If the springs remain horizontal and the mass is made to rotate at an angular speed of half$rev/s$, then the value of spring force $F_1$ and $F_2$ (in newton) are.

• A. $3.3,3.3$
• B. $3.3,6.6$
• C. $6.6,3.3$
• D. $0,0$

Answer 1

The correct option is B $3.3,6.6$

FBD of the system is


Since, springs are parallel
$k_{eq}=k_1+k_2=300N/m$

and let the deflection in both springs be $x$ (say)

Here, $r=$ radius of circular path$=$ Natural length of spring $=0.5m$
and $\omega =\frac{1}{2}rev/s=\pi rad/s$

The spring force acts towards the centre, therefore it provides the necessary centripetal force.
i.e $k_{eq}x=mr{\omega }^2$
$\Rightarrow 300\times x=2\times 0.5\times (\pi {)}^2$
$\therefore x=\frac{{\pi }^2}{300}=0.0329m=3.3cm$

Value of spring forces $F_1$ and $F_2$ are given by:
$F_1=k_{1}x=100\times 3.3\times {10}^{-2}=3.3N$
$F_2=k_{2}x=200\times 3.3\times {10}^{-2}=6.6N$