Transverse Wave Velocity (speed)

  • Thread starter paul11273
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Prof gave us this equation and question:

The equation of a transverse wave traveling in a rope is:
y=10sin*pi(0.01x-2.00t)

He said that x and y are expressed in cm, and time in seconds.

We are to:
A. find the amplitude, frequency(Hz), velocity(cm) and wavelength(cm) of the wave.
B. find the maximum transverse speed of a particle in the rope.

I am not sure how to answer these, because I do not know if the amplitude of 10 is in meters or cm. Also, does k=rad/m or rad/cm? I am confused since he specified that x and y are in cm. Does this make everything given in the equation cm? Or is there some standard that keeps them as meters?

I know that you cannot read my professors mind. I am hoping there is some sort of standard that applies here, or someones experience can win out over my prof's vagueness. Of course, this is due the next time that class meets (Thursday), so no time to clarify with him.

Any help would be great.
 

Answers and Replies

  • #2
Tide
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I think you have to assume cgs units from the statement of the problem.
 
  • #3
HallsofIvy
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"I know that you cannot read my professors mind."

It's not necessary to read his mind, he told you everything out loud!

"He said that x and y are expressed in cm, and time in seconds."

x and y are in cm. sin( whatever) has no units, its just a function. If y is in cm. then the right hand side has to be in cm. The amplitude, 10, has to be in cm. because you are told that y is.
Similarly, the argument of the sin function, 0.01x-2.00t, has to be in radians. Since you know that x is in cm, that 0.01 must have units of radians/cm. Since you know that t is in sec., that 2.00 must have units of radians/sec.
 
  • #4
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Halls: Thanks, and would you mind having a look at my solutions?

Part A:
Amplitude = 10 cm from the equation
Frequency:
f=w/2pi (where I am using w as omega)
= 2.00/2pi
= 1/pi ~= .318 Hz
Velocity:
v=w/k
=2/.01
= 200 cm/s

Wavelength:
lambda=v/f
=200/.318 ~= 629cm

Part B: To find the max velocity of the transverse wave.
This is a little tricky to me.

Given the partial derivative WRT t found earlier we have

Partial dy/dt= -20pi*cos(pi(.01x - 2.00t))

Now I understand that the maximum will occur when the trig portion is equal to 1. That leaves me with only the -20pi

The text gives v,max of the transverse as v,max=wA
I believe that all I needed to do here is multiply -20*pi to get the v,max.

However, I am not understanding how the units work out to give cm/s.
It should (-20rad/s)*pi*(1cm).
What happens to the radian? I never quite grasped this type of situation from previous physics classes. It is some type of psuedo unit that drops?

Please advise. I am trying to get it. Thanks.
 
  • #5
nrqed
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paul11273 said:
Halls: Thanks, and would you mind having a look at my solutions?

Part A:
Amplitude = 10 cm from the equation
Frequency:
f=w/2pi (where I am using w as omega)
= 2.00/2pi
= 1/pi ~= .318 Hz
Velocity:
v=w/k
=2/.01
= 200 cm/s

Wavelength:
lambda=v/f
=200/.318 ~= 629cm

Part B: To find the max velocity of the transverse wave.
This is a little tricky to me.

Given the partial derivative WRT t found earlier we have

Partial dy/dt= -20pi*cos(pi(.01x - 2.00t))

Now I understand that the maximum will occur when the trig portion is equal to 1. That leaves me with only the -20pi

The text gives v,max of the transverse as v,max=wA
I believe that all I needed to do here is multiply -20*pi to get the v,max.

However, I am not understanding how the units work out to give cm/s.
It should (-20rad/s)*pi*(1cm).
What happens to the radian? I never quite grasped this type of situation from previous physics classes. It is some type of psuedo unit that drops?

Please advise. I am trying to get it. Thanks.

You made several times the same error. Remember that there is a [itex] \pi [/itex] inside your sine function. For example, [itex] \omega = 2.00 \pi rad/sec [/itex] and not 2.00 rad/sec. Your speed is actually correct because you forgot two factors of [itex] \pi [/itex] which actually cancel out. So you got it right by chance. But your wavelength is wrong. By the way, you can also calculate the wavelength using [itex] \lambda = 2 \pi/k[/itex].

You get the correct max transverse speed (but if you really want the speed, you must drop the minus sign).


As for the radian, it is indeed a bit confusing. It's not really a unit, it's a kind of "pseudo" unit as you say :tongue2: that drops out. Basically, the radians are just there to remind us that the angles are given in radians and not in degrees, but angles don't really have units, so you often see the radians disappearing mysteriously. So you were absolutely right.

Pat
 
  • #6
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Thanks Pat. I will recalculate my answers in the morning and post. You are correct, I completely forgot about the pi.
 
  • #7
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OK, so my corrected answers are:

Amplitude = 10cm
Frequency = 1Hz
Velocity = 200 cm/s
Wavelength = 200 cm
Max Transverse speed = 20*pi cm/s ~= 62.8 cm/s

I feel good about this type of problem. Thanks for the help and pointing out my errors.

P.S. If the above is still not correct, please let me know :tongue2:
 
  • #8
The radians are taken care of because they are a dimensionless unit, so when they are multiplied by the unit of cm-with dimension length-the radinas are simply canceled out.
 

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