Distance traveled by a particle in a transverse wave

[greg_rack]

Homework Statement

A transverse wave traveling through a medium has a frequency of 5.0 Hz, a wavelength of
4.0 cm and an amplitude of 3.0 cm.
What is the total distance traveled by a particle of the medium in one minute?

Relevant Equations

$v=\lambda f$
$f=\frac{1}{T}$

Taken into account the transverse nature of the wave, I deduce the particle must move of harmonic motion from maximum amplitude $A$ to negative maximum amplitude $-A$.
The period $T=\frac{1}{f}$ is equal to the time in which a particle travels a distance $d=3\cdot A$. I then approximated the mean vertical velocity of the particle $v_{y}=\frac{3\cdot A}{T}$.
Then, multiplying this result per 60 seconds, I should find the distance traveled in a minute... but the result I get is wrong.
There must be a problem with the mean velocity...

 

[LCSphysicist]

Homework Statement:: A transverse wave traveling through a medium has a frequency of 5.0 Hz, a wavelength of
4.0 cm and an amplitude of 3.0 cm.
What is the total distance traveled by a particle of the medium in one minute?
Relevant Equations:: $v=\lambda f$
$f=\frac{1}{T}$

Taken into account the transverse nature of the wave, I deduce the particle must move of harmonic motion from maximum amplitude $A$ to negative maximum amplitude $-A$.
The period $T=\frac{1}{f}$ is equal to the time in which a particle travels a distance $d=3\cdot A$. I then approximated the mean vertical velocity of the particle $v_{y}=\frac{3\cdot A}{T}$.
Then, multiplying this result per 60 seconds, I should find the distance traveled in a minute... but the result I get is wrong.
There must be a problem with the mean velocity...

"The period $T=\frac{1}{f}$ is equal to the time in which a particle travels a distance $d=3\cdot A$. I then approximated the mean vertical velocity of the particle $v_{y}=\frac{3\cdot A}{T}$"
No, it raise up A, returns again to equilibrium position by traveling A, it goes down A, and returns, again, traveling A! It is not 3, it is four.

 

[greg_rack]

"The period $T=\frac{1}{f}$ is equal to the time in which a particle travels a distance $d=3\cdot A$. I then approximated the mean vertical velocity of the particle $v_{y}=\frac{3\cdot A}{T}$"
No, it raise up A, returns again to equilibrium position by traveling A, it goes down A, and returns, again, traveling A! It is not 3, it is four.

Thank you very much, that's right