How would you go about canceling out the effect of a radio wave by emitting another wave of same frequency, wavelength and amplitude
[Physics] Cancel out radio waves
electromagnetic-radiationwaves
Related Solutions
In amplitude modulation, the frequency of the carrier wave is constant. The frequency spectrum of an AM signal includes sidebands, but those aren't the carrier wave. In your second figure, the carrier wave is the black line. You'll note that the amplitude changes; it increases and decreases in accordance with the modulation, however the frequency of this wave does not change. That is the essence of amplitude modulation. For a carrier wave of constant frequency, the information is encoded in the amplitude of the signal. The presence of sidebands does not imply the frequency is non-constant, merely that the overall signal is not a single pure frequency. The carrier still remains the same throughout and the frequency of the AM wave is connotatively the same as the frequency of the carrier.
Consider the incoming electric field of the radio waves. This field is a superposition of all broadcasts from stations near your receiver. The job of the receiver is to pick out one of these transmissions and turn it into sound.
AM radio
Now consider an AM radio station transmitter. Suppose the sound wave that station wants to transmit is represented by a function of time $m(t)$ where here $m$ is for "message". Note that $m(t)$ includes all information about the sound, i.e. it includes frequency, amplitude... everything. In an AM transmitter, we use a circuit to multiply $m(t)$ by a sinusoid, creating the transmitted signal $$s(t) = m(t) \cos(\Omega t)$$ where here $s$ stands for "signal" and $\Omega$ is called the "carrier frequency". Here we see the reason for the term Amplitude Modulation (AM): the message is a modulation of the amplitude of the carrier wave.
You can use trig identities or Fourier analysis to see that the spectral content of $s(t)$ is in the range $\Omega \pm \delta \omega$ where $\delta \omega$ is the highest frequency in $m(t)$. The carrier frequency $\Omega$ might be in the tens of MHz range. On the other hand, the actual message $m(t)$ would absolutely never have any frequencies above around 20 kHz because that's the upper range of human hearing. In real life, $m(t)$ doesn't use up the full 20 kHz; useful speech and music don't need our full hearing range.
So now we see that the transmitted signal $s(t)$ is contained within some relatively narrow bandwidth, i.e. maybe a 10 kHz band centered at 10 MHz. Therefore, a tuned circuit with a $Q$ of around 1,000 and centered at $\Omega$ picks up $s(t)$ but mostly nothing else.$^{[a]}$ Of course, we also have to enforce that the various stations' carrier frequencies are separated by more than their $\delta \omega$'s so that nobody's transmissions overlap with anyone else's.
So, the output of our tuned circuit is roughly just $s(t)$! I say "roughly" because our tuned circuit isn't perfect, so we might pick up a bit of stuff from other transmissions, but since it's farther away from the center of our tuned circuit the amplitude is suppressed. Then, we just put the signal through a rectifier and a low pass filter so that the carrier oscillations are gone and we only get $m(t)$. That's it! Now we have the original sound message and we can put it into a speaker. We don't have to think about amplitude and frequency separately: we have the entire original sound waveform.
$[a]$: $Q$ is the center frequency divided by the bandwidth, so $$Q = 10 \text{MHz} / 10 \text{kHz} = 1,000 \, .$$
Best Answer
It is possible to cancel sound waves by detecting incoming sound waves and then generating another sound wave of the same frequency and direction but opposite phase. Because electrical signals can travel much faster than sound, the sound wave that cancels the incoming sound wave can be generated a bit downstream from a microphone, just in time for the incoming sound wave to be canceled. In other words, an electrical signal can be sent ahead of the sound wave to "instruct" a sound generator to produce the cancelling sound wave when the incoming wave arrives, and that makes cancellation possible.
However, it is not possible to send a signal "ahead" of an incoming radio wave, because nothing can travel faster than a radio wave (or any other electromagnetic wave), so active cancellation the way you envision can't be done.
There is a kind of complicated situation in which it is "sort of" possible: if the incoming radio wave is moving very slowly because of the medium it is passing through. Paraffin, for example, slows the speed of radio frequencies to about 2/3 the speed of light. If the incoming radio waves had to pass through a block of paraffin hundreds of meters thick, but were detected as they entered and an optical signal were sent ahead via a "tunnel" through the paraffin, the optical signal could in principle get to the far side of the paraffin before the radio waves arrive, in time to "instruct" a radio source to generate the right radio wave to cancel the incoming wave just as it arrives. I know that's not what you had in mind, but it's about as close to what you want as is physically possible.