Question

A 297 g block is connected to a light spring with spring constant 4.34 N/m, and displaced 7.45 cm from equilibrium. It is then released and allowed to oscillate in simple harmonic motion. What is the maximum acceleration of the block

Answer 1

Answer:

x = A sin ω t       describes the displacement of the particle

v = A ω cos ω t

a = -A ω^2 sin ω t        

a (max) = -A ω^2     is the max acceleration (- can be ignored here)

ω = (K/ m)^1/2        for SHM

F = - K x^2     restoring force of spring

K = 4.34 / .0745^2 = 782 N / m

ω = (782 / .297)^1/2 = 51.3 / sec

a (max) = .0745 * 782 / .297 = 196 m / s^2