Question

A spring stretches 50mm when a mass of 0.15kg is hung from it (g=data sheet). The spring is extended a further 15mm and realeased. Its Time Period and Maximum acceleration are?

Answer 1

here,

mass of block, m = 0.15 kg

net extension, x = 50 + 15 = 65 mm = 0.065 m

From newton second law : SUM(F) = 0
spring force = force duw to gravity
k * x = m * g

spring constant, k = mg/x
spring constant, k = (0.15 * 9.81)/(0.065)
spring constant, k = 22.64 N/m

time peroid, T = (2*pi) * sqrt(m/k)
time peroid, T = (2*pi) * sqrt(0.15/22.64)
time peroid, T = 0.511 s

Max acceleration, a = x * (2*pi/T)^2
Max acceleration, a = 0.065 * (2*pi/0.511)^2
Max acceleration, a = 9.877 m/s^2