Question

Two ideal springs of same make (the springs differ in their lengths only) have been suspended from points $A$ and $B$ such that their free ends $C$ and $D$ are at same horizontal level. A massless rod $PQ$ is attached to the ends of the springs. The block of mass $m$ is attached to the rod at point $R$. The rod remains horizontal in equilibrium. Now the block is pulled down and released. It performs vertical oscillations with time period, $T=2\pi \sqrt{\frac{m}{3k}}$ where $k$ is the force constant of the longer spring. Choose the correct statements.

• A. The effective spring constant in $3k$
• B. For linear oscillations (without wobbling) to take place, $RC:RD=2:1$
• C. The ratio of lengths of springs $AC$ and $BD$ is $3:1$
• D. The ratio of lengths of springs $AC$ and $BD$ is $2:1$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The effective spring constant in $3k$
B For linear oscillations (without wobbling) to take place, $RC:RD=2:1$
D The ratio of lengths of springs $AC$ and $BD$ is $2:1$

The time period of vertical oscillations of the block is,
$T=2\pi \sqrt{\frac{m}{3k}}$
This means that the effective force constant is $3k=k_{AC}+k_{BD}$ [Springs $AC$ and $BD$ are in parallel]
$3k=k+k_{BD}$
$k_{BD}=2k$.
Spring $BD$ must have a force constant of $2k$.

Rod $PQ$ is massless. Net force and torque about $R$ on it must be zero.
$\Rightarrow RC\left[kx\right]=RD\left[2kx\right]$
$\Rightarrow \frac{RC}{RD}=\frac{2}{1}$

Force constant of spring $\propto$ $\frac{1}{\text{Length of spring}}$

$k_{AC}{L}_{AC}=k_{BD}{L}_{BD}$
$k{L}_{AC}=2k{L}_{BD}$
$L_{AC}=2L_{BD}$
$\therefore$ Length of $AC$ must be double that of spring $BD$