An amplitude-modulated wave is represented by, $$C_m(t) = 10(1 + 0.2\cos 12560t)\sin(111 \times 10^4 t)$$ V. The modulating frequency in kHz will be _________

We start by recalling the standard mathematical description of a sinusoidal amplitude-modulated (AM) signal. In general, the instantaneous value of an AM wave can be written as

$$C(t)=A_c\bigl(1+m\cos\omega_m t\bigr)\sin\omega_c t$$

where

$$A_c$$ is the carrier amplitude,

$$m$$ is the modulation index,

$$\omega_m$$ (in rad s−1) is the angular frequency of the modulating (message) signal, and

$$\omega_c$$ (in rad s−1) is the angular frequency of the carrier.

Comparing the given expression

$$C_m(t)=10\bigl(1+0.2\cos 12560\,t\bigr)\sin\!\bigl(111\times10^4\,t\bigr)\;{\rm V}$$

with the standard form, we make term-by-term identification:

$$A_c = 10,$$

$$m = 0.2,$$

$$\omega_m = 12560\ \text{rad s}^{-1},$$

$$\omega_c = 111\times10^4\ \text{rad s}^{-1}.$$

Our objective is to find the numerical value of the modulating frequency $$f_m$$ in kilohertz. The relationship between angular frequency and ordinary (linear) frequency is

$$f = \frac{\omega}{2\pi}.$$

Applying this formula to the modulating signal, we have

$$f_m = \frac{\omega_m}{2\pi} = \frac{12560}{2\pi}\ \text{Hz}.$$

Carrying out the division, we proceed step by step:

First compute the denominator:

$$2\pi \approx 6.2832.$$

Now divide:

$$f_m = \frac{12560}{6.2832}\ \text{Hz}.$$

Evaluating the quotient gives

$$f_m \approx 1999.8\ \text{Hz}.$$

Since 1999.8 Hz is effectively 2000 Hz when rounded to a reasonable number of significant figures, we may write

$$f_m \approx 2000\ \text{Hz}.$$

Finally, converting hertz to kilohertz (recall that $$1\ \text{kHz}=1000\ \text{Hz}$$) yields

$$f_m = \frac{2000\ \text{Hz}}{1000} = 2\ \text{kHz}.$$

Hence, the correct answer is Option 2.