Question

A vertical spring-mass system has a mass of 0.25 kg and an initial deflection of 0.5 cm. Then what will be the value of spring stiffness

Answer 1

F = Mg = Kx

0.25 * 10 = K * 0.5/100

k= 500

Answer 2

Answer:

K = 500 Nm²

Explanation:

Given: Mass of the vertical spring = 0.25 kg, Initial deflection = 0.5 cm

To Find: Value of spring stiffness.

Procedure:

${\bf{\boxed{\bf{kx = mg}}}}$

$\implies \sf k = 0.5^{-2}m = 0.25 \times 10$

$\implies \sf k= 2.5$

$\implies \sf k = \dfrac{25}{5 \times 10^{-2}}$

$\implies \sf k = \dfrac{5}{10^{-2}}$

$\implies \sf k = 5 \times 10^2$

$\implies \sf k = 5 \times 100$

$\implies \bf k = 500 Nm^2$

• A vertical spring-mass system is a system of springs connected by different masses which is meant to go in simple harmonic motion in "vertical" direction.