Question

A spring is hanging from the ceiling. Attaching a 500 g physics book to the spring causes it to stretch 20 cm in order to come to equilibrium. a. What is the spring constant? b. From equilibrium, the book is pulled down 10 cm and released. What is the period of oscillation? c. What is the book’s maximum speed? At what position or positions does it have this speed?

Answer 1

Answer:

a. 25 N/m

b. 0.8886 s

c. 0.707 m/s

d. At the equilibrium point

Explanation:

m = 500 g = 0.5 kg

L = 20 cm = 0.2 m

A = 10 cm = 0.1 m

a. Let g = 10 m/s2, then the gravity of the 0.5 kg book acting on the spring is

F = mg = 0.5*10 = 5 N

If the spring is stretched L = 0.2 m under 5N load, then the spring constant k is:

k = F/l = 5 / 0.2 = 25 N/m

b. We can treat this as simple harmonic motion with magnitude A = 0.1 cm. The period of this motion is

$T = 2\pi \sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{0.5}{25}} = 0.8886 s$

c. The book maximum speed:

$\omega A = \sqrt{\frac{k}{m}}A = \sqrt{\frac{25}{0.5}}0.1 = 0.707 m/s$

d. Due to the law of energy conservation, the maximum speed would occur at the equilibrium point. This is where the potential energy, elastic energy is 0 and the kinetic energy is greatest.