ine csampe Example 43 A mencuey themomeer baving s time coss

ine csampe Example 4.3 A mencuey themomeer baving s time cossiam ok &1 in placed in u sempenstere hath n l 00-F nd lowed to comne to equilibrium wists toil its, erige temperature of ,\'NFF with an amplitude of 7-F. If the frequency reading as a function of chox. What is the phase l4gT In of she symbols weed in this chapter 0.1 min syelesis From Eq. (428), the amplitude of the response and the phase angle are calculated: thus

Solution

In the given problem, term f represents freqency which is equal to (10/pi) in cycles/min that is Hz.

Now, omega (w) is nothing but angular velocity in rad/min.

1 Hz = 1 cycle/sec = 6.28 rad/sec = 2*pi rad/sec

Therefore, (10/pi) rad/sec = (2*pi) * (10/pi) rad/sec = 20 rad/sec

and similarly (10/pi) rad/min = ((2*pi) * (10/pi) rad/min = 20 rad/min

This is just a coversion of cycle/min to rad/min.