Question

Two identical springs, each have k = 20.0 N/m. A 1.00 kg is connected to
them as shown. Calculation the period of motion of the oscillatory system.
Neglect frictional forces.​

Answer 1

Answer:

Explanation:

You find a spring in the laboratory. When you hang 100 grams at the end of the spring

it stretches 10 cm. You pull the 100 gram mass 6 cm from its equilibrium position and

let it go at t = 0. Find an equation for the position of the mass as a function of time t.

Lets first find the period of the oscillations, then we can obtain an equation for

the motion. The period T = 2π

q

m/k. The mass m is 0.1 Kg. To find k, we use the

fact that 100 grams causes the spring to stretch an additional 10 cm. Since F = k∆x,

we have

mg = k∆x

0.1(9.8) = k(0.1)

k = 9.8 N/m

The period of the motion is therefore T = 2π

q

0.1/9.8 ≈ 0.635 sec. At t = 0 the

mass is at its maximum distance from the origin. Thus, x(t) = 0.6 cos(2πt/T). Using

T = 0.635 sec gives

x(t) = 0.6 cos(2πt/0.635)

x(t) ≈ 0.6 cos(9.9t)

The cosine function is the appropriate one, since at t = 0 the mass is at its maximum

distance from equilibrium.