Water Wave Physics Problem: Calculating New Wavelength in Refraction Zone

AI Thread Summary
In a ripple tank experiment, students need to calculate the new wavelength of water waves as they refract into shallower water, where the speed decreases from 4.0 cm/s to 3.0 cm/s. The frequency remains constant during refraction, and the relationship between wave speed, frequency, and wavelength is described by the equation v = λν. The students initially confused frequency with period but clarified that frequency is measured in Hertz, while period is in seconds. After correcting their calculations, they confirmed that the new wavelength can be determined using the new speed and the constant frequency. The discussion concludes with the students gaining clarity on the concepts and calculations involved.
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Homework Statement


In a ripple tank experiment, students generate water waves at a speed of 4.0 cm/s and a wavelength of 0.5 cm. If the waves are refracted into shallower water where their speed decreases to 3.0 cm/s, what is their new wavelength?


Homework Equations





The Attempt at a Solution


i have no idea about anything here!please help?!
 
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Well, one of the first places to start is with the equation relating the velocity of a wave, its frequency and its wavelength. Next, when the waves refract into shallower water and begin traveling slower, does this change the waves' wavelength, frequency, or both? Please provide your reasoning for these answers.
 


ooo so i would find the force using f=v/ λ, then plug that force and the new speed into find the new wavelength. do i have to change everything from cm to m?
 


Alright, backtracking it looks like you know the equation:

v = \lambda \nu

Where v is the velocity, \lambda is the wavelength, and \nu is the frequency (be careful not to get the force and frequency mixed up, they are two very different things).

Now, it also looks like you realized that when the waves refract, this does nothing to the frequency. Using, this frequency you can calculate exactly what you need! I think that you can answer your question about unit conversions if you give it enough thought (look at the mathematical expression you're dealing with and make sure units cancel).
 


ok, so if the units for frequency is Hz and that's what I am finding, it gives me T because the meters cancel. so would i go 1/T to get the frequency then plug that number into get the actual new wavelength?
 


Why don't you write out what you're thinking mathematically and I'll check that ;-). It's a bit difficult to follow what you wrote.
 


F= v/ λ
F=0.04m/s/0.005m
F=8s----->frequency isn't measured in s, its measured in Hz. so..
F=1/T
F=1/8s
f=0.125Hz
λ=v/f
λ= 0.03m/s/0.125hz
=0.24m

or do i just keep the f in seconds and do this instead.
F= v/ λ
F=0.04m/s/0.005m
F=8
λ=v/f
λ= 0.03m/s/8
λ=3.8 x 10^-3m
 


Okay, it looks like you're confused in a couple of places . . .

First, you need to be careful with your math, one of your manipulations with the units essentially amounts to:

\frac{m}{s}*\frac{1}{m} = s

Hopefully you can understand why that's incorrect. Second, the equation

\nu = \frac{v}{\lambda}

Gives you the frequency of the wave, not the period! It's very important to get these concepts correct. Try again and we'll see where it goes from there.
 


from what your telling me.. I am not sure what is wrong with the first way? because i did change the frequency into the period
 
  • #10


Here's what I mean more explicitly . . .

\nu = \frac{v}{\lambda} = \frac{4 \mathrm{cm/s}}{0.5 \mathrm{cm}} = 8 \frac{1}{\mathrm{s}} = 8 \mathrm{Hz} \neq 8 \mathrm{s}
 
  • #11


i knwo that 8 hz doesn't equal 8 seconds, which is why i went f=1/t, f=1/8 which gives me Hz does it not?
 
  • #12


Okay, you're still confused. The frequency is measured in Hertz and the period is measured in seconds. The equation that you're using gives you the frequency, not the period. It can, however, be easily modified to obtain the period. To check that you have the right frequency, find the product of the wavelength and the frequency (if this doesn't give you the speed, you've done something wrong). Does this clear things up? As an aside, is this algebraic manipulation okay with you?

\frac{a}{b} * \frac{1}{a} = b
 
  • #13


ok, so when i tried solving the problem, what part did i do wrong.

and yes, i understand that. the a's are cancelling each other out
 
  • #14


Well, the part that you did wrong is that after you calculated the frequency, you went on and calculated the period and called that the frequency. Seriously, find the product of the wavelength and frequency and if it doesn't give you the velocity then your frequency is incorrect. I'll do this for you using the values you calculated:

0.125 \mathrm{Hz} * 0.5 \mathrm{cm} = 0.0625 \frac{\mathrm{cm}}{\mathrm{s}}

You should note that this isn't the velocity specified in the problem. Hopefully you wouldn't have understood that "algebraic" manipulation because it was completely fallacious. It should be equal to 1/b, not b.
 
  • #15


ok so 8hz * 0.5cm does work out to the right speed. (4). in that case what is wrong with the second way i tried.
 
  • #16


What's wrong with the second way is that it still seems that you're confused about the distinction between the period and frequency since you said:

or do i just keep the f in seconds

Try to present your solution again, this time using the correct names for concepts.
 
  • #17


k just look at this part of the equation.
1)F= v/ λ
2)F=0.04m/s/0.005m
3)F=8
----> m/s *1/m. so since the m's cancel out, were left with 1/s which is Hz right? because if u plugged a number in fr seconds it would be like the equation, f=1/T.

4)Λ=v/f
5)Λ= 0.03m/s/8
6)=3.8 x 10-3
in step 5 I am plugging in the new speed and the found frequency
 
  • #18


Yes, your work looks correct. You're correct about the meters cancelling and leaving you with 1/s which is the same unit as Hz. Also, you don't need to convert everything into meters - leaving everything in terms of centimeters works just fine. Good job!
 
  • #19


phew! thank you so much for your time. :)
 
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