Question

When a 13.8-kg block is attached to the free end of a vertically hanging spring, he spring stretches by 21 cm. Wjat would the mass of the block have to be in order for the spring to stretch by 31.0 cm?

Answer 1

Given
mass, m= 13.8 kg
x = 21 cm = 0.21 m
a) since F = kx
mg = kx
13.8 kg * 9.8 m/s^2 = k * 0.21 m
k = 644 N/m
b) period , T = 2π√m/k

   T = 2 * 3.14 *√13.8 kg/644 N/m

    T = 0.91 s

c)Frequency , f = 1/T = 1/0.91s

             f = 1.09 Hz

d)The amplitude is the maximum displacement from the equilibrium

position . so A = 31cm

e)Maximum speed , v = Aω = A(2π/T)

                                 = (0.31 m) (2 *3.14)/0.91s

                         v = 2.13 m/s