Questions on superpositioning.

@jazzjeppe
I just want to know if I’m getting it right for me.

Isn’t it something to do with magnetic fields or something? I honestly don’t have a clue, but I am also interested to know.

@jazzjeppe
It may be the waves off different parts of the metal are interfering destructively but it says “Some of the output of the transmitter (by the wire shown )directly to the receiver and is then compared with the reflected signal.” What makes me think otherwise.

The wavelength is 30×10^-3 m which is 30mm
The reciever picks up the reflected sigal.
If you look at the wavelength of a wave which is also called one period, you can see the wave does a full cycle in one wavelength::: top of peak, through a trough, to top of peak. Or max intensity through min intensity, back to max.

You move the reflector 7.5mm closer.
Divide 30mm (wavelength) by 7.5mm (movement of the metal reflector) and you get ¼ wavelength. Because the signal is being reflected it loses 7.5mm from its movement along its wavelength on the way to sheet and coming back.

So ¼ wavelength x 2 = ½

If you receieved max intensity first time, then cut a half wavelength out of the waves path – you recieve at min intensity, which is a half wavelength along the waves path

It is all to do with interference. At its original position, the two components of the wave interfere constructively causing a maximum, and when it is moved they interfere destructively resulting in a minimum. If these two points are the minimum and maximum, you can conclude that the wavelength of the signal is 30mm (confirming the given information), as wavelength is measured crest to crest, and you need to double the distance because it is reflected, and so travels the distance twice.

@FireMadeFlesh
What two waves though? so you have oone wave that reflects off the metal and the other
?

@Siblinings “Some of the output of the transmitter (by the wire shown) directly to the receiver and is then compared with the reflected signal.”

What you send to the receiver via the wire can be seen as a second wave, because its information must be in a similar state to the emitted wave if they are to be compared (superpositioned) to each other.

Only problem with the interference solution is that if initially you have a maximum at the reciever (sum of two peaks), and then you change the phase difference between the two by ½ period, you end up with destructive interference and a sum signal of zero, not a minimum. (taking minimum to mean the sum of two troughs). But then if minimum means zero signal, it works for me (and two troughs summing would be a negative max).

@jahono Good point. I can’t think of any other solution though, because we haven’t been given any information on the properties of the detector. My guess is that it is a poorly written question.