Question

2. A 2kg body attached to a spring undergoes a SHM of amplitude 0.4m and period

0.Sts. a) What maximum force acts on a body? b) Find the acceleration of the body
at a distance of 0.2m for m the equilibrium position.​

Answer 1

a) 12.8 N

b) 3.2 m/s²

Step-by-step explanation:

I'm guessing the period is 0.5π s.

Period of a spring in simple harmonic motion is:

T = 2π √(m/k)

Given T = 0.5π and m = 2 kg:

0.5π = 2π √(2/k)

0.25 = √(2/k)

0.0625 = 2/k

k = 32

The spring constant is 32 N/m, and the maximum displacement is 0.4 m. The maximum force can be found with Hooke's law:

F = kx

F = (32 N/m) (0.4 m)

F = 12.8 N

The acceleration can be found with Newton's second law:

∑F = ma

kx = ma

(32 N/m) (0.2 m) = (2 kg) a

a = 3.2 m/s²