Question

A block is tied within two springs, each having spring constant equal to $k$. Initially the springs are in their natural length and horizontal as shown in fig, the block is released from rest. The springs are ideal, acceleration due to gravity is $g$ downwards. Air resistance is to be neglect. The natural length of the spring is $\left({\ell }_0\right)$ . the decrease in height of the block till it reaches equilibrium is $\sqrt{3}{\ell }_0$

• A. The mass of the block is $\frac{\sqrt{3}{\ell }_{0}k}{g}$
• B. Velocity of the block becomes zero at the equilibrium position.
• C. Maximum speed of the block in motion when it is released from the shown horizontal position is $\sqrt{\frac{4{\ell }_{0}g}{\sqrt{3}}}$
• D. The acceleration of the block, just after cutting any one of the strings is less than $g$.

Answer 1

The correct option is A The mass of the block is $\frac{\sqrt{3}{\ell }_{0}k}{g}$

(Refer to Image)

When a block is tied within two springs, each having spring constant equal to= $k$

Initially the springs are in their natural length and horizontal

$u=0$ (block is released with rest)

Acceleration due to gravity= $g$ (downward)

Air resistance is to be neglected, the natural length of the spring is $l_o$ .

The decrease in height of the block till it reaches equilibrium= $\sqrt{3}{l}_o$

Using hook's law, it states that,

$F\propto x$

$F=Kx$

or, $ma=Kx$ (where using Newton's 2nd law $F=ma$)

or, $m=\frac{Kx}{a}=\frac{Kx}{g}$ $a=g=$gravitational force

now, $x$= Given= $\sqrt{3}{l}_o$

The mass will become $=\frac{K\sqrt{s}{l}_o}{g}$

$=\frac{\sqrt{3}{l}_{o}K}{g}$ .