# 物理好難阿 請懂的進來替看看2

1 A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant k and mass m. If the damping constant has a value b1, the amplitude is A1 when the driving angular frequency equals (k/m)1/2. In terms of A1, what is the amplitude F max, if the damping constant is (a) 3b1 and (b) b1/2

2 An square object with mass m is constructed of four identical uniform thin sticks, each of length L, attached together. This object is hung on a hook at its upper corner (Fig. 13.37.). If it is rotated slightly to the left and then released, at what frequency will it swing back and forth?

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Problem-1

General Equation for One DOF (Degree of Freedom) Vibration System:

m*x” cx’ kx = F0*sin(w*t)

It is a second order ordinary differential equation. Its solution can be found in many textbooks as:

x(t) = (F0/k)*sin(w*t-theta) * (1/D)

where

theta = atan(2zeta(w/wn)/(1-(w/wn)^2),

D = √{[1-(w/wn0^2]^2+[2*zeta*(w/wn)]^2},

wn = (1/2*pi)*√(k/m) and

zeta = c / (2*√(k*m)).

Thus when the driving angular frequency equals √ (k/m) (i.e. driving force is in system resonant frequency), D becomes

D = 2*zeta

It means the Fmax ~∝ 1/zeta ∝ 1/c. (~ means: proportional)

So if the damping constant is 3b1, its amplitude = A1 * (b1/3b1) = A1/3.

Similarly, when damping constant is b1/2, its amplitude = A1*(b1/0.5b1) = 2*A1.

Problem-2

Since the square object with mass m is suspended by four identical uniform thin sticks, this object always keeps horizontal during swinging motion. In other words, it won't have any rigid body rotational motion with respect to its own center of gravity. Thus, it can be treated as a pendulum of length L with a concentrated mass m at the end. So its resonance frequency is

f = (1/2*pi)* √(g/L) (in Hz).