Question

A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to its equilibrium position. This stretches the spring by a length x. If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case

• A. x
• B. 2x
• C. 3x
• D. x/2

Answer 1

Answer: (B)

2x

Answer 2

When spring is gradually lowered to it's equilibrium position

kx = mg ∴ $x = \frac { m g } { k }$.

When spring is allowed to fall suddenly it oscillates about it's mean position

Let y is the amplitude of vibration then at lower extreme, by the conservation of energy

⇒ $\frac { 1 } { 2 } k y ^ { 2 } = m g y$ ⇒ $y = \frac { 2 m g } { k }$ = 2x.