Max Displacement of Particle at x=0.5cm

[lha08]

Homework Statement

What is the maximum displacement of a particle at x=0.5 cm
The equation for the resulting standing wave: y(x,t)= 4sin(1.26x)cos(50.27t) cm.

Homework Equations

The Attempt at a Solution

Like the answer tells me that 4sin(1.26*0.5)=ymax but I'm not sure why we would disregard the cos(50.27t) part...does it have to do with the antinode and node idea because in my book it says that antinodes occurs wherever sink=(plus minus)1...any clarification would be much appreciated

 

[berkeman]

Homework Statement

What is the maximum displacement of a particle at x=0.5 cm
The equation for the resulting standing wave: y(x,t)= 4sin(1.26x)cos(50.27t) cm.

Homework Equations

The Attempt at a Solution

Like the answer tells me that 4sin(1.26*0.5)=ymax but I'm not sure why we would disregard the cos(50.27t) part...does it have to do with the antinode and node idea because in my book it says that antinodes occurs wherever sink=(plus minus)1...any clarification would be much appreciated

As t varies, what is the maximum value of cos(50.27t)?

 

[lha08]

As t varies, what is the maximum value of cos(50.27t)?

Well i suppose that we're not given t so do we ignore it?

 

[Redbelly98]

No, we don't ignore t. We allow t to have any value possible.

And so the maximum value of cos(50.27t) is ?

 

[lha08]

No, we don't ignore t. We allow t to have any value possible.

And so the maximum value of cos(50.27t) is ?

ummm the amplitude?

 

[Ouabache]

ummm the amplitude?

The amplitude will vary depending on what you choose for t.
How about plugging in a range of values for t in cos(50.27t) and plot your results? Does it have a maximum?

 

[CompuChip]

The amplitude will vary depending on what you choose for t.
How about plugging in a range of values for t in cos(50.27t) and plot your results? Does it have a maximum?

Well, actually lha is right in this one... the amplitude is the maximum value attained over all t and therefore does not depend on t. The point we're trying to get across, lha, is that you want the maximum value of
y(t) = 4sin(1.26*0.5)cos(50.27t).

The way we like to do that is by forgetting about the 4 sin(...) part in front, and just looking at the cosine. Where the cosine is at its maximum, so is y(t) (for example, if the cosine takes the value 10 somewhere and is smaller everywhere, then if you multiply it by 4 sin(1.26*0.5) you will get something which takes the value 4*sin(1.26*0.5)*10 somewhere and is smaller everywhere).

 

[Ouabache]

Well, actually lha is right in this one... the amplitude is the maximum value attained over all t and therefore does not depend on t.

I believe we are both on the same page CC, perhaps we have a slight difference in terminology. My meaning of amplitude is the height of the plot, when graphing y(t) = 4sin(1.26*0.5)cos(50.27t), on Cartesian coordinates.

The amplitude will vary depending on what you choose for t.

In the present example, after you substituted x the 4sin(1.26*0.5) is a constant and becomes a scaling factor. As you vary t, the height (amplitude) of the plot will vary as a function of cos(50.27t), but does have a maximum at a certain value. Once you have determined this value, multiplying by the scaling factor does give you the maximum displacement of the particle.