You're not telling us where this pops up but my some magic I believe to know that you're talking about the frequency response of a continuous-time linear time-invariant system. It is defined as the Fourier transform of the system's impulse response $h(t)$:

$$H(j\omega)=\int_{-\infty}^{\infty}h(t)e^{-j\omega t}dt$$

Its practical relevance is that it shows the frequency dependence of the system's input-output relation. If $X(j\omega)$ is the Fourier transform of the input signal, and $Y(j\omega)$ is the Fourier transform of the output signal, then
$$H(j\omega)=\frac{Y(j\omega)}{X(j\omega)}$$

For a sinusoidal input signal $x(t)=\sin (\omega_0 t)$ the output signal $y(t)$ is given by

$$y(t)=|H(j\omega_0)|\sin (\omega_0 t + \text{arg}\{H(j\omega_0\})$$

where $\text{arg}\{H(j\omega_0\})$ is the phase of the complex function $H(j\omega)$ at frequency $\omega_0$. So you can see that $|H(j\omega)|$ represents the amplification of the system at frequency $\omega$. From above relations it is also obvious that the output signal cannot contain any frequencies that were not present in the input signal. $H(j\omega)$ can only reduce or amplify frequency components of the input signal. This is how (linear time-invariant) filtering works.