Finding the Maximum Positive Transverse Velocity of a Transverse Wave

[sugz]

Homework Statement

A transverse wave is described by y=0.02 sin(30x-400t). Determine the first positive x-coordinate (x>0) for which the transverse velocity of that point in the medium is at its maximum positive value when t=0. All variables are in SI units.

a) 5.2 cm b) 10.4cm c) 15.7 cm d.20.9 e) none of the above.

Homework Equations

The Attempt at a Solution

I found the derivative of the position equation to get the equation of velocity, at t=0 which is v = 0.6 cos(30x). I equated it to max velocity which is 0.6, therefore getting 1=cos(30x). arcos(1) is 2pi, which I later divided by 30 to get 0.209.
Where did I make a mistake?

 

[Nathanael]

y=0.02 sin(30x-400t)
...
v = 0.6 cos(30x)

Double check your differentiation. Is this the correct dy/dt?

Edit: p.s. arccos(1) is 0 but that's not relevant.

 

[sugz]

I assumed we would be differentiating in terms of x, how do we know which variable we are differentiating in terms of?

 

[Nathanael]

I assumed we would be differentiating in terms of x, how do we know which variable we are differentiating in terms of?

The definition of velocity is the derivative with respect to time.

 

[sugz]

Oh right, my mistake. Also, even if I were to differentiate with respect to time, the equation obtained is y= 8 cos(30x-400t) and I plug in the value of t=0, getting the equation of y= 8 cos(30x). The position when the velocity is at maximum positive velocity, which is 8. Therefore you get the equation 8=8cos(30x). I will still get the same results, no?

 

[Nathanael]

Oh right, my mistake. Also, even if I were to differentiate with respect to time, the equation obtained is y= 8 cos(30x-400t)

That's not quite the right equation. You're missing a factor of negative 1, aren't you?

 

[sugz]

Oh right, now I see where I made a mistake! Thank you so much!