Simple harmonic motion

Question

Consider a spring-mass system on a horizontal frictionless surface. The system has a spring constant of 250N/m and a mass of 0.9kg. The system is released with an initial compression of the spring of 0.13m from the equilibrium point and an initial speed of the mass of 4m/s. What is the maximum stretch of the spring during the simple harmonic motion?

Group of answer choices

0.25 m

0.38 m

0.13 m

0.27 m

Answer 1

This can be solved by conservation of energy.

Let

k = 250 N/m be the spring constant,

m = 0.9 kg be the mass attached to the spring,

d = 0.13 m be initial displacement,

v = 4 m/s be initial velocity,

x be the maximum stretch be to calculated.

When the spring reaches the maximum stretch x, the spring energy will be kx²/2.

That energy must be equal the initial energy, which has two components:

the initial spring energy kd²/2, and

the initial kinetic energy mv²/s

kx²/2 = kd²/2 + mv²/2

x = √(kd² + mv²)/k = √(d² + mv²/k)

= √(0.12² + 0.9×4²/250) = 0.27 m