A block is attached to two unstretched springs and with spring constants and , respectively (see fig. I). The other ends are attached to identical supports and not attached to the walls. The springs and supports have negligible mass. There is no fri

Question

A block $(B)$ is attached to two unstretched springs $S_{1}$ and $S_{2}$ with spring constants $k$ and $4 k$ , respectively (see fig. I). The other ends are attached to identical supports $M_{1}$ and $M_{2}$ not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block $B$ is displaced towards wall 1 by a small distance $x$ (figure II) and released. The block returns and moves a maximum distance $y$ towards wall 2. Displacements $x$ and $y$ are measured with respect to the equilibrium position of the block $B$ . The ratio $y / x$ is [2008]

Smart Achievers · Accepted Answer

(3)

Here when the block B is displaced towards wall 1, only spring $S_{1}$ is compressed and $S_{2}$ is in its natural state as the other end of $S_{2}$ is free.

Therefore the energy stored in the system $= \frac{1}{2} k_{1} x^{2}$

When the block is released, it will come back to the equilibrium position, gain momentum, overshoot the equilibrium position and move towards wall 2. As this happens, the spring $S_{1}$ comes to its natural length and $S_{2}$ gets compressed. The P.E. stored in the spring $S_{1}$ gets stored as the P.E. of spring $y / x$ when the block $B$ reaches its extreme position after compressing $S_{2}$ by $y$ . It is because no friction anywhere.

So, energy is conserved

$∴$ $\frac{1}{2} k_{1} x^{2} = \frac{1}{2} k_{2} y^{2} \Rightarrow \frac{1}{2} k x^{2} = \frac{1}{2} \times 4 k y^{2}$

$\Rightarrow x^{2} = 4 y^{2}$ $∴$ $\frac{y}{x} = \frac{1}{2}$

Solution

1 4

2 2

3 1/2

4 1/4

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