Question

Figure shows a block of mass $m$ attached to a spring of force constant $k$ and connected to ground by two string. In relaxed state natural length of the spring is $l$. In the situation shown in figure, find the tension in the strings (1) and (2).

• A. $T_1=\frac{1}{4}[kl-2\text{mg}],T_2=\frac{\sqrt{3}}{4}[kl-2\text{mg}]$
• B. $T_1=\sqrt{3}[kl-2\text{mg}],T_2=[kl-2\text{mg}]$
• C. $T_1=\frac{\sqrt{3}}{4}[kl-2\text{mg}],T_2=\frac{1}{4}[kl-2\text{mg}]$
• D. $T_1=[kl-2\text{mg}],T_2=\sqrt{3}[kl-2\text{mg}]$

Answer 1

The correct option is C $T_1=\frac{\sqrt{3}}{4}[kl-2\text{mg}],T_2=\frac{1}{4}[kl-2\text{mg}]$

As the natural length of spring is $l$, and in the situation shown in figure its length is $\frac{3l}{2}$. Thus the spring is stretched by a distance $\frac{l}{2}$ hence it exerts a restoring force on block $k\left(\frac{l}{2}\right)$ upward as shown in figure, which shows also the tensions acting on the block along the directions of the strings. As the block is in equilibrium, we can balance all the forces acting on it along horizontal and vertical directions.

Along horizontal direction
$T_1\text{sin}{30}^{\circ }=T_2\text{sin}{60}^{\circ }$
or $T_1=\sqrt{3}{T}_2$.......(1)
Along vertical direction
$k\left(\frac{l}{2}\right)=mg+T_1\text{cos}{30}^{\circ }+T_2\text{cos}{60}^{\circ }$
or $\sqrt{3}{T}_1+T_2=kl-2\text{mg}$.......(2)
Substituting value of $T_1$ from equation (1) in equation (2) we get,
$4T_2=kl-2\text{mg}$
$T_2=\frac{1}{4}(kl-2\text{mg})$
From equation (1), we have
$T_1=\frac{\sqrt{3}}{4}(kl-2\text{mg})$