Math question with physics equation.

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In summary: FT=1N, FT=2N, use any µ you like. Which frequencies do you get?In summary, the frequencies are not directly proportional.
  • #1
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Homework Statement


Hello guys I am having problems with wave and frequency problems. I know the equation use to find frequency are ∫=λ/v and
∫=1/2L([itex]\sqrt{FT/μ}[/itex])

So my question is: if the length and μ are kept the same will FT be directly proportional? Will frequency increase as the tension increase?

*v=[itex]\sqrt{FT}[/itex]
I have the same question about velocity when the μ is kept constant.

Homework Equations


v=[itex]\sqrt{FT}[/itex]
∫=1/2L([itex]\sqrt{FT/μ}[/itex])

The Attempt at a Solution


I said when the velocity increases as the tension increases. I know that numbers in the numerator are directly proportional to what you are trying to find but the square root is throwing me off.
 
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  • #2
Please specify the physics problem you are considering - I guess waves on a string with mass density µ.

So my question is: if the length and μ are kept the same will FT be directly proportional?
It does not make sense to ask "is [variable] proportional"? "Proportional" is a relation between two variables. The frequency and tension are not proportional to each other due to the square root - just plug in numbers to verify that.
##v=\sqrt{F_T}##
That is not right.

I said when the velocity increases as the tension increases.
Right.
 
  • #3
mfb said:
Please specify the physics problem you are considering - I guess waves on a string with mass density µ.

It does not make sense to ask "is [variable] proportional"? "Proportional" is a relation between two variables. The frequency and tension are not proportional to each other due to the square root - just plug in numbers to verify that.
That is not right.

Right.
Thank you for the reply. I am having problems with the equations and general common sense question. Let's say an object on a string experiences Simple Harmonic Motion. The strings have the same linear densities and lengths. String one has FT while string two has 2FT. What can be said about their frequencies?

From the equation ∫=1/2L ([itex]\sqrt{FT/μ}[/itex])

By removing the constants I am left with f=[itex]\sqrt{FT}[/itex]
Q: By increasing the string by 2FT will the frequency increase by 2?
 
  • #4
jvdamdshdt said:
By removing the constants I am left with f=[itex]\sqrt{FT}[/itex]
No, you get [itex]f=c\sqrt{F_T}[/itex] with some constant c.
20=5*4 does not imply 20=4 just because 5 is constant!

Alternatively, you can write [itex]f \propto \sqrt{F_T}[/itex] ("f is proportional to the square root of FT").

Q: By increasing the string by 2FT will the frequency increase by 2?
Did you plug in some numbers?
FT=1N, FT=2N, use any µ you like. Which frequencies do you get?
 
  • #5


As a scientist, it is important to understand the relationship between variables in equations and how they affect each other. In this case, the equations you have provided are related to wave and frequency problems in physics. The first equation, v=λ/f, describes the relationship between velocity (v), wavelength (λ), and frequency (f). The second equation, v=√(FT/μ), describes the relationship between velocity (v), tension (T), and mass per unit length (μ) in a string or rope.

To answer your question, if the length and μ are kept constant, then yes, FT will be directly proportional. This means that as tension (T) increases, frequency (f) will also increase. This can be seen in the first equation, as an increase in velocity (v) leads to a decrease in wavelength (λ), causing an increase in frequency (f).

Similarly, if μ is kept constant, then an increase in tension (T) will also result in an increase in velocity (v). This is because an increase in tension will cause an increase in the speed of the wave, as described by the second equation.

It is important to note that the square root in the second equation does not affect the direct proportionality between T and v. This is because the square root is applied to both T and μ, so their relationship remains the same.

In summary, yes, frequency and velocity will both increase as tension increases, as long as the other variables (length and μ) are held constant. I hope this helps you better understand these equations and their relationships. Keep up the good work with your wave and frequency problems!
 

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