Question

A spring is hung from the ceiling. A 0.473 -kg block is then attached to the free end of the spring. When released from rest, the block drops 0.109 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

Answer 1

Answer:

a)

85.05 N/m

b)

179.81 rad/s

Explanation:

a)

k = spring constant of the spring

m = mass of the block = 0.473 kg

x = stretch caused in the spring = 0.109 m

h = height dropped by the block = 0.109 m

Using conservation of energy

Spring potential energy gained by the spring = Potential energy lost by the block

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) k x = mg

(0.5) k (0.109) = (0.473) (9.8)

k = 85.05 N/m

b)

angular frequency is given as

$w = \sqrt{\frac{k}{m}}$

$w = \sqrt{\frac{85.05}{0.473}}$

$w$ = 179.81 rad/s