Step 1: Form the resultant wave.
The superposition of the two waves gives: \[ y = y_1 + y_2 = 10 \cos(kx) \cos(\omega t) \] At \(x = 0\): \[ y = 10 \cos(\omega t) \]
Step 2: Determine when velocity is zero.
Velocity of the particle: \[ v = \frac{dy}{dt} = -10 \omega \sin(\omega t) \] The velocity becomes zero when \(\sin(\omega t) = 0\). \[ \omega t = n\pi \Rightarrow t_n = \frac{n\pi}{\omega} \]
Step 3: Use given time intervals.
Velocity is zero at \(t = 0,\ 0.25,\ 0.5\). The time difference between consecutive zeros: \[ t_{n+1} - t_n = \frac{\pi}{\omega} = 0.25 \] \[ \omega = \frac{\pi}{0.25} = 4\pi \]
Step 4: Calculate frequency.
\[ f = \frac{\omega}{2\pi} = \frac{4\pi}{2\pi} = 2\, \text{Hz} \]
Step 5: Conclusion.
Hence, the minimum frequency of the waves is \(2\, \text{Hz}\).
Using a variable frequency ac voltage source the maximum current measured in the given LCR circuit is 50 mA for V = 5 sin (100t) The values of L and R are shown in the figure. The capacitance of the capacitor (C) used is_______ µF.

