Discussion
Consider the following linear second order differential equation
\[ \frac{d^{2}y}{dt^{2}} + \omega^{2}y = 0, \]
where \(\omega\) is a positive constant. The boundary conditions are \(\left.\frac{dy}{dt}\right|_{t=0} = 1\), and \(y(t = 0) = \frac{1}{2}\).
Which of the following options gives the value of \(y(t = \frac{\pi}{2\omega})\)?
Answer: A
The value is \(\frac{1}{\omega}\).
The general solution of \(\ddot{y} + \omega^2 y = 0\) is \(y = A\cos\omega t + B\sin\omega t\). The two conditions fix the constants:
\[ \begin{aligned} y(0) &= A = \tfrac{1}{2} \\ \dot{y}(0) &= B\omega = 1, \quad B = \tfrac{1}{\omega} \end{aligned} \]
At \(t = \frac{\pi}{2\omega}\), \(\omega t = \frac{\pi}{2}\), so the cosine term vanishes and the sine term is at its maximum:
\[ \begin{aligned} y\left(\tfrac{\pi}{2\omega}\right) &= \tfrac{1}{2}\cos\tfrac{\pi}{2} + \tfrac{1}{\omega}\sin\tfrac{\pi}{2} \\ &= \tfrac{1}{\omega} \end{aligned} \]
0 is what you get by keeping only the cosine term, and 1 by forgetting to divide the initial slope by \(\omega\).
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development