Difference between displacement and amplitude

How do I calculate one given the other (and time period)
What is the difference between a time vs amplitude sinusoidal graph and a displacement vs
distance traveled sinusoidal graph? I'm confused.
The question is Physics: Electrons, Waves and Photons Monday 6 June 2011 Question 5,

Reply 1

uberteknik

Original post by Primus2x

No difference other than context, semantics and brevity:

Displacement is associated with a visible physical movement, e.g. swing of a pendulum, vibrations on the string of a guitar.

Amplitude is normally associated with conceptual waves like em propagation, sound waves, pressure waves etc.

However you can also refer to the amplitude of a string's vibration which is the same as saying displacement distance between a node and anti-node.

In reality, both mean the same thing but the context determines which to use.

(edited 10 years ago)

Reply 2

Stonebridge

In the context of the question referred to

http://www.ocr.org.uk/Images/65871-question-paper-unit-g482-electrons-waves-and-photons.pdf

there is a distinction between displacement (of a point on the wave from its equilibrium position) and the amplitude of that oscillation.
Amplitude is the maximum displacement of the point. It's a in the formula

y = a sin wt
y is displacement
a is amplitude

That question has about 4 sections and about 10 subsections.

Perhaps the OP could specify which parts or questions he would like help with, otherwise it's difficult to answer meaningfully when only presented with a vague query and no link to the exam question.

(edited 10 years ago)

Reply 3

Primus2x

Original post by Stonebridge

In the context of the question referred to

http://www.ocr.org.uk/Images/65871-question-paper-unit-g482-electrons-waves-and-photons.pdf

there is a distinction between displacement (of a point on the wave from its equilibrium position) and the amplitude of that oscillation.
Amplitude is the maximum displacement of the point. It's a in the formula

x = a sin wt
x is displacement
a is amplitude

That question has about 4 sections and about 10 subsections.

Perhaps the OP could specify which parts or questions he would like help with, otherwise it's difficult to answer meaningfully when only presented with a vague query and no link to the exam question.

I got 0.6m for the wavelength in part a i, I'm not sure whether I'm correct with part a ii, the wave moves twice (as shown with the dotted curve) and the period for one wave is 0.75 so I multiplied 2 and 0.75 to show the time period is 1.5, is this right?
But the main part I am stuck on is part b, mainly the first part with the calculations.

Reply 4

Stonebridge

Original post by Primus2x

Yes, fig 5.1 shows the wave as it actually appears in space but frozen at a particular moment. The wavelength is 0.6m as this can be clearly measured directly from the graph by taking the distance between any two equivalent points. It's the length of one complete wave cycle.
The next part is correct. The wave has moved a half of a wavelength in the given time of 0.75ms so it would take 2 x 0.75 = 1.5ms to move a whole wavelength. This is the period of the wave.

b)
As the wave moves from left to right the two points P and Q will move up and down. They will both reach the maximum displacement as the crest (and trough) of the wave moves by.
So they will both have the same amplitude (maximum possible displacement) but at any time their displacements won't necessarily be the same. They will reach their max displacements at different times. As the wave is moving from left to right, it gets to P before Q, so P will reach its greatest height before Q.

Difference between displacement and amplitude

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