Question

Problem 3: Find the natural frequency of the system shown in Figure 3. Problem 4: In the mechanical system shown in Figure 4, assume that the rod is massless, perfectly rigid, and pivoted at point P. The displacement x is measured from the equilibrium position. Assuming that x is small, that the weight mg at the end of the rod is 5 N, and that the spring constant k is 400 N/m, find the natural frequency of the system. 2a 2 ng ky Figure 3: Mechanical System Figure 4: Mechanical System

Answer 1

Problem 3.

From the figure, it is evident that all the springs K₁ , K₂ and K₃ are connected in parallel, hence equivalent spring constant :

K_eq = K₁ +K₂ + K₃

Now,

$m\ddot{x} = -K_{eq}x$

$\ddot{x} = -\frac{K_{eq}}{m}x$

$\ddot{x} = -\omega ^2x$

$\omega = \sqrt{\frac{K_{eq}}{m}} = \sqrt{\frac{K_{1}+K_{2}+K_{3}}{m}}$

Problem 4.

By applying energy method,

KE + PE = Constant

$\frac{d}{dt}(KE + PE) = 0$

$KE = \frac{1}{2}m\dot{x}^2$

$PE = \frac{1}{2}k(x/2)^2$

$\frac{d}{dt}(\frac{1}{2}m\dot{x}^2 + \frac{1}{2}k(x/2))^2) = 0$

$m\dot{x}\ddot{x} + \frac{1}{4}kx\dot{x} = 0$

$\dot{x}(m\ddot{x} + \frac{1}{4}kx) = 0$

$\dot{x}\neq 0$

$\ddot{x} = -\frac{k}{4m}x$

$\omega =\sqrt{\frac{k}{4m}}$

mg = 5 N

Taking g = 10 m/s².

m = 5/10 = 0.5 kg

k = 400 N/m

$\omega =\sqrt{\frac{400}{4*0.5}} = \sqrt{200} = 14.142 Hz$