Question

A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position

Answer 1

t = 9.52 s

Step-by-step explanation:

This is an oscillatory motion exercise, in which the angular velocity is

w =
`\sqrt{ (k)/(m) }`

Let's use hooke's law to find the spring constant, let's write the equilibrium equation

F_e - W = 0

F_e = W

k x = m g

k =
`(m g)/(x)`

k = 0.545 9.8 /0.0356

k = 150 N / m

now the angular velocity is related to the period

W = 2π / T

we substitute

4π² T² = k /m

T = 4pi²
`\sqrt{ (m)/(k) }`

we substitute

T = 4 pi²
`\sqrt{ (0.545)/(150) }`

T = 2.38 s

therefore for the spring to oscillate 4 complete periods the time is

t = 4 T

t = 4 2.38

t = 9.52 s