Question

In the given figure, the block is attached with a system of three ideal springs $A,B$ and $C$. The block is displaced by a small distance $x$ from its equlibrium position vertically downwards and released. $T$ represents the time period of small vertical oscillations of the block (pulleys are ideal). Choose the correct alternative.

• A. $T=2\pi \sqrt{\frac{11m}{2k}}$
• B. the deformation of the spring $A$ is $\frac{2}{11}$ times the displacement of the block.
• C. the deformation of the spring $C$ is $\frac{1}{11}$ times the displacement of the block.
• D. the deformation of the spring $B$ is $\frac{4}{11}$ times the displacement of the block.

Answer 1

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

the deformation of the spring $B$ is $\frac{4}{11}$ times the displacement of the block.

Let the elongations in springs $A,B$ and $C$ be $x_1,x_2$ and $x_3$ respectively.

Let $a$ be the acceleration of block, When a block is displaced by $x$.

Using constraint relations we can write that,

$x=x_1+2x_2+x_3.....\left(1\right)$

Let $\text{T}$ be the tension developed in string due to displacement of block.

Thus, from the figure it is evident that,

$T=2k{x}_3$ , $2T=k{x}_2$ and $T=k{x}_1$

Using these in $\left(1\right)$ we get,

$x=\frac{T}{k}+\frac{2\left(2T\right)}{k}+\frac{T}{2k}$

$\Rightarrow x=\frac{11T}{k}$

$\Rightarrow x=\frac{11ma}{k}$

$\Rightarrow a=\left(\frac{k}{11m}\right)x$

Comparing this with $|a|=|{\omega }^{2}x|$ we get,

Time period $T=2\pi \sqrt{\frac{11m}{2k}}$

Also, $x_2=4x_3....\left(2\right)$ and $x_2=2x_1......\left(3\right)$

Using $\left(2\right)$ and $\left(3\right)$ in $\left(1\right)$

$x_1=\frac{2}{11}x;x_2=\frac{4}{11}x;x_3=\frac{1}{11}x$

Hence, all the options are the correct alternatives.