Question

A horizontal plank of mass m and length L is pivoted at one end. The plank’s other end is supported
by a spring of force constant k. The plank is displaced by a
small angle

from its horizontal equilibrium position and
released. (a) Show that it moves with simple harmonic motion
with an angular frequency

= 3k/m . (b) Evaluate the
frequency if the mass is 5.00 kg and the spring has a force
constant of 100 N/m

Answer 1

To develop this problem we will be guided by the graph that fits the description of the problem. Performing sum of torques we will find the displacement of the body after the displacement of a small angle, this would be

$\sum \tau = 0-mg(\frac{L}{2})+ kx_0 L$

Here

$x_0$ = the equilibrium compression

After displacement by a small angle

$\sum \tau = -mg(\frac{L}{2})+kxL$

$\sum \tau = -mg(\frac{L}{2})+k(x_0-L\theta)L$

$\sum \tau = -k\theta L^2$

At the same time we have that the angular torque can be defined as,

$\tau = I \alpha$

Here,

$\text{Moment of Inertia} = I = \frac{1}{3} mL^2$

$\text{Angular acceleration} = \alpha = \frac{d^2\theta }{dt^2}$

Replacing this value we have that

$\tau = \frac{1}{3} mL^2 (\frac{d^2\theta}{dt^2})$

Replacing the previous value for the angular torque found,

$-k\theta L^2 = \frac{1}{3} mL^2(\frac{d^2\theta}{dt^2})$

$\frac{d^2\theta}{dt^2} = \frac{-3k}{m} \theta$

Angular acceleration is opposite in direction and proportional to the displacement so we have simple harmonic motion with

$\omega^2 = \frac{3k}{m}$

$\omega = \sqrt{\frac{3k}{m}}$

PART B) Evaluating for m = 5kg and k = 100N/m we have that

$\omega = \sqrt{\frac{3k}{m}}$

$\omega = \sqrt{\frac{3(100)}{5}}$

$\omega = 7.745rad/s$