Question

In the arrangement shown in figure, pulleys are small and light and the springs are ideal. $K_1=25{\pi }^2\frac{N}{m},K_2=2K_1,K_3=3K_1$ and $K_4=4K_1$ are the force constants of the springs. Calculate the period of small vertical oscillations (in $\text{s}$) of the block of mass $m=3\text{kg}$

Answer 1

In static equilibrium of the block, tension in the string is exactly equal to its weight. Let a vertically downward force $F$ be applied on the block to pull it downwards. Equilibrium is again restored when tension in the string is increased by the same amount $F$. Hence, total tension in the string becomes equal to $(mg+F)$.
Strings are further elongated due to the extra tension $F$. Due to this extra tension $F$ in the strings, tension in each spring increases by $2F$ [since pulleys are light and massless, force in spring $=2\times$ Tension in string.
Hence, increase in elongation of the springs are $\frac{2F}{K_1},\frac{2F}{K_2},\frac{2F}{K_3},\frac{2F}{K_4}$ respectively.

Accoding to the geometry of the arrangement, downward displacement of the block from its equilibrium position is
$y=2(\frac{2F}{K_1}+\frac{2F}{K_2}+\frac{2F}{K_3}+\frac{2F}{K_4})........\left(i\right)$
If the block is released now, it starts to accelerate upwards due to extra tension $F$ in the strings. It means restoring force on the block is equal to $F$.
From Eq. (I),
$F=\frac{y}{4(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4})}$
Restoring acceleration of the blocks
$a=\frac{F}{m}=\frac{y}{4m(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4})}$
Since acceleration of the block is resotring and is directly proportional to displacment $y$, the block performs SHM.
Time period, $T=2\pi \sqrt{\frac{\text{displacement}}{\text{acceleration}}}$
$T=2\pi \sqrt{4m(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4})}$
$T=4\pi \sqrt{m(\frac{1}{K_1}+\frac{1}{K_2}+\frac{1}{K_3}+\frac{1}{K_4})}$
$T=4\pi \sqrt{m\times \frac{25}{12K_1}}$
After substituting the values, we get $T=2\text{s}$.