Speed of Propagation vs. Particle Speed....please help?

Question

Speed of Propagation vs. Particle Speed.

The equation
y(x,t)=Acos2(pi)f((x/v)-t)

may be written as

y(x,t)=Acos[(2pi/lambda)(x-vt)]

Use the last expression for to find an expression for the transverse velocity of a particle in the string on which the wave travels

Please help....i've been stuck on this forever...=)

Answer 1

genericman1998 has good ideas however here is another way.

Since in a transverse wave the particle displacement is perpendicular to the direction of wave propagation we can safely say that the time it take from the particle to go from the upper most position to the lowest most position and back up it period T. Therefore

V=S/t where
S- distance traveled during time t
t - same as period T
S= A + A + A + A = 4A (I'm not being facetious.)

V=4A/T since T=1/f
finally
V=4Af


In contrast the wave propagation speed is
v=Lf (L is for lambda :-) )
L- wavelength
f - frequency

So what is the wavelength? The wavelength is the path the particle has to travel during a full period. It is a circular motion where our circle has a radius A.
So L= 2piA and

v= 2piAf or
v=wA where
w- is angular frequency w=2 pi f

Have fun

Answer 2

My answer assumes you've had some calculus. Sorry if that isn't the case, but I don't know any other way.

Recall that velocity is the derivative of position with respect to time. So, "transverse" velocity is the derivative of y (the "transverse" position) with respect to time.

Differentiate the expression y(x,t) = ... with respect to t. Treat "x" as a constant when you do this.

Good luck, hope this helps.