Question

Interactive Solution 10.23 presents a model for solving this problem A vertical spring (spring constant =180 N/m) is mounted on the floor. A 0.570-kg block is placed on top of the spring and pushed down to start it oscillating in simple harmonic motion. The block is not attached to the spring. (a) Obtain the frequency (in Hz) of the motion. (b) Determine the amplitude at which the block will lose contact with the spring.

Answer 1

(a) k = 180 N/m

mass m = 0.570 kg

So, frequency of the motion -

f = (1 / 2?) * ?(k / m)

= (1 / 2?) * ?(180 / 0.570) = (1 / 2?) * ?316 = 17.8 / 6.28 = 2.83 Hz

(b) w = 2?*f = 2*3.14*2.83 = 17.8 rad /s

In the case of simple harmonic motion -

x(t) = A * cos(w * t)

then velocity -

v(t) = - A * w * sin(w * t)

again acceleration -

a(t) = - A * w^2 * cos(w * t),

where A is the maximum amplitude.

In the limiting case, when the block is on the verge of losing contact with the spring -

a(t) = g = 9.81m/s^2.

so -

a(x) = - w^2 * x = 9.81 m/s^2

=> 9.81 = (17.8)^2 * x. And solving for x, the displacement,

=> x = 9.81 /315.8 = 0.031 m = 3.10 m

So, the requisite amplitude = 3.10 cm