Transverse Wave

  • Thread starter Husker70
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Homework Statement


A transverse wave on a string is described by the wave function
y = (.120m) sin(pie/8 + 4pie(t))

(a) Determine the transverse speed and acceleration of an element of the
string at t=0.200s for the point on the string located at 1.60m
(b) What are the wavelength, period, and speed of propagation of this wave?


Homework Equations


Part (a)
Velocity = -wA cos(kx-wt)
Acceleration = -w^2A sin(kx-wt)

Part B
Wavelength = k = 2pie/lambda
Period = 1/f
Speed = f(lambda)
w = 2pief
f = w/2pie = 2.0 rad/s

The Attempt at a Solution



Part (a)
V = -(4pie) (.120m)cos(pie/8 + 2.51) I don't know what to do from here
A = -(4pie)^2 (.120m) sin(pie/8 + 2.51) I don't know what to do from here

Part (b)
Wavelength = pie/8 = 2pie/lambda = 16m
Period = 1/2.0 s^-1 = .5s
Speed f(lambda) = (2.0s^-1)(16m) = 32m/s

If anyone can help I would appreciate it.
Thanks,
Kevin

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
Redbelly98
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(.120m) sin(pie/8 + 4pie(t))

Hi,

Is that supposed to be
(.120m) sin((x)pie/8 + 4pie(t))
?
(An "x" is needed for this to be a wave.)

V = -(4pie) (.120m)cos(pie/8 + 2.51) I don't know what to do from here
A = -(4pie)^2 (.120m) sin(pie/8 + 2.51) I don't know what to do from here

Just plug the numbers into your calculator, using radian mode for the sine calculation. And again, please check if it should really be
(x)pie/8

Wavelength = pie/8 = 2pie/lambda = 16m

You got the right answer. However, what you wrote here is bad "math grammar". It says (among other things) that
Wavelength = pie/8
which is wrong. pie/8 is actually "k", which as you know is different than wavelength.

A better way to write this out would be something like this:

Wavelength:
k = pi/8 m^-1 = 2pi/lambda
lambda = 2pi / (pi/8 m^-1) = 16 m


Your period and speed calculations look correct.
 

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