Solution Found!
Solved: (a) Take first and second derivatives with respect
Chapter 30, Problem 42E(choose chapter or problem)
(a) Take first and second derivatives with respect to time of \(q\) given in Eq. (30.28), and show that it is a solution of Eq. (30.27). (b) At \(t=0\) the switch shown in Fig. 30.17 is thrown so that it connects points \(d\) and \(a\); at this time, \(q=Q\) and \(i=d q / d t=0\).Show that the constants and in Eq. (30.28) are given by
\(\tan \phi=-\frac{R}{2 L \sqrt{(1 / L C)-\left(R^{2} 4 L^{2}\right)}} \text { and } A=\frac{Q}{\cos \phi}\)
Equation Transcription:
Text Transcription:
q
t=0
d
a
q=Q
i=dq/dt=0
phi
tanphi=-R over 2L sqrt (1/LC)-(R^2/4L^2) and A=Q over cosphi
Questions & Answers
QUESTION:
(a) Take first and second derivatives with respect to time of \(q\) given in Eq. (30.28), and show that it is a solution of Eq. (30.27). (b) At \(t=0\) the switch shown in Fig. 30.17 is thrown so that it connects points \(d\) and \(a\); at this time, \(q=Q\) and \(i=d q / d t=0\).Show that the constants and in Eq. (30.28) are given by
\(\tan \phi=-\frac{R}{2 L \sqrt{(1 / L C)-\left(R^{2} 4 L^{2}\right)}} \text { and } A=\frac{Q}{\cos \phi}\)
Equation Transcription:
Text Transcription:
q
t=0
d
a
q=Q
i=dq/dt=0
phi
tanphi=-R over 2L sqrt (1/LC)-(R^2/4L^2) and A=Q over cosphi
ANSWER:Introduction
We have to show that the expression of charge given at equation (30.28) is the solution of the equation (30.27). Then, using the boundary condition, we have to calculate the quantity
and
in the expression of charge.
Step 1
The expression of charge given at the equation (30.28) is given by
Now differentiating the above equation with respect to the time we have
Now differentiating the above equation again with respect to time we have
The equation (30.27) is given by
Now putting the values in the righthand side of the above equation we have
Hence it is proved that the expression of charge given in the equation (30.28) is a solution of equation (30.27).