To the free end of spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is d. If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is: (force constant, k )

Question

To the free end of spring hanging from a rigid support, a block of mass m is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is d. If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is: (force constant, k ) (A) (m g/k) (B) 2 d (C) (m g/3 k) (D) 4 d

Tardigrade Admin · Accepted Answer

To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is

x_{1}

, then
Work done by weight of the block

=

potential energy stored in spring

m g x = \frac{1}{2} k x^{2}

∴ x = 2 \frac{m g}{k} = 2 d

(∵ d = \frac{m g}{k})

$\frac{m g}{k}$

$2 d$

$\frac{m g}{3 k}$

$4 d$

To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is $x_{1}$ , then
Work done by weight of the block $=$ potential energy stored in spring
$m g x = \frac{1}{2} k x^{2}$
$∴ x = 2 \frac{m g}{k} = 2 d$
$(∵ d = \frac{m g}{k})$

kmg​

2d

3kmg​

4d

To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is x1​, then Work done by weight of the block = potential energy stored in spring mgx=21​kx2 ∴x=2kmg​=2d (∵d=kmg​)

$\frac{m g}{k}$

$2 d$

$\frac{m g}{3 k}$

$4 d$

To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is $x_{1}$ , then
Work done by weight of the block $=$ potential energy stored in spring
$m g x = \frac{1}{2} k x^{2}$
$∴ x = 2 \frac{m g}{k} = 2 d$
$(∵ d = \frac{m g}{k})$