Question

If x1=3sinωt and x2 = 4cosωt then

• A. $\frac{x_1}{x_2}$ is independent of t
• B. Average value of $<x_1^2 + x_2^2>$ from t = 0 to t = $\frac{2\pi}{\omega}$ is 12.5
• C. $(\frac{x_1}{3})^2 + (\frac{x_2}{4})^2 = 1$
• D. Average value of $$ from t = 0 to t = $\frac{2\pi}{\omega}$ is zero

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

1. Option (A):

   $$\frac{x_1}{x_2} = \frac{3\sin\omega t}{4\cos\omega t} = \frac{3}{4}\tan\omega t \quad (\text{depends on } t)$$

   So, (A) is false.

2. Option (B):

   $$x_1^2 + x_2^2 = 9\sin^2\omega t + 16\cos^2\omega t$$

   Over one period, the averages are:

   $$\langle\sin^2\omega t\rangle = \langle\cos^2\omega t\rangle = \frac{1}{2}$$

   Thus,

   $$\langle x_1^2+x_2^2\rangle = 9\left(\frac{1}{2}\right) + 16\left(\frac{1}{2}\right) = \frac{9+16}{2} = \frac{25}{2} = 12.5$$

   So, (B) is true.

3. Option (C):

   $$\left(\frac{x_1}{3}\right)^2+\left(\frac{x_2}{4}\right)^2 = \sin^2\omega t + \cos^2\omega t = 1$$

   So, (C) is true.

4. Option (D):

   $$x_1x_2 = 3\sin\omega t \times 4\cos\omega t = 12\sin\omega t\cos\omega t = 6\sin2\omega t$$

   The average of $\sin2\omega t$ over one full period is zero. So, (D) is true.