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Chapter 1

Q. 1.3

The response of a system is given by

x\left(t\right) =0.003 \cos \left(30t \right)+0.004\sin \left(30t \right)m          (a)

Determine (a) the amplitude of motion, (b) the period of motion, (c) the frequency in Hz, (d) the frequency in rad/s, (e) the frequency in rpm, (f ) the phase angle, and (g) the response in the form of Equation (1.12)

Step-by-Step

Verified Solution

(a) The amplitude is given by Equation (1.13) which results in

X=\sqrt{0.003^{2}+0.004^{2}  } m =0.005m          (b)

(b) The period of motion is

T=\frac{2\pi }{30}s+0.209s      (c)

(c) The frequency in hertz is

f=\frac{1}{T} =\frac{1}{0.209 s}=4.77 Hz          (d)

(d) The frequency in rad/s is 

\omega =2\pi f=30  rad/s          (e)

(e) The frequency in revolutions per minute is

\omega =\left(20\frac{rad}{s} \right)\left(\frac{1rev}{2\pi  rad} \right)\left(\frac{60 s}{1 min} \right)  = 191.0  rpm          (f)

(f) The phase angle is

\phi =\tan ^{-1}\left( \frac{0.003}{0.004}\right) =0.643 rad          (g)

(g) Written in the form of Equation (1.12), the response is

x\left(t\right) =0.005\sin \left(30 t+0.643 \right)m          (h)

 

x\left(t\right) =X\sin \left(\omega t+\phi \right)          (1.12)

X=\sqrt{A^{2}+B^{ 2} }          (1.13)