Trick Questions?

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DDS

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Trick Questions??

A stretched wire vibrates in its fundamental mode at a frequency of 378 Hz. What would be the fundamental frequency if the wire were half as long, its diameter were doubled, and its tension were increased three-fold?

AND

A beam containing light of wavelengths λ1 and λ2 is incident on a set of parallel slits. In the interference pattern, the fourth bright line of the λ1 light occurs at the same position as the sixth bright line of the λ2 light. If λ1 is known to be 539 nm, what is the value of λ2?


for question 1) i know that the relationship describing fundamental mode is:

F1= 1/2L * sqrt Ts/U

however i cant get over the fact that your not given any real numbers to determine the new fundamental frequency. I have tried t osub those numbers into the equation and manipulate them but i keep getting 1134 Hz

for question 2)

we know that lamda 1-4 is = lamda2-6 however i figured that
lamda 2 would be equal to lamda one but its not.


any hints for these questions???
 

Doc Al

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DDS said:
for question 1) i know that the relationship describing fundamental mode is:

F1= 1/2L * sqrt Ts/U

however i cant get over the fact that your not given any real numbers to determine the new fundamental frequency. I have tried t osub those numbers into the equation and manipulate them but i keep getting 1134 Hz
You know that
[tex]F_1 = \frac{1}{2 L} \sqrt{\frac{T}{\mu}}[/tex]
To find F_2, substitute the new length, tension, and mu. For example, replace L with L/2, etc. Then compare with the original expression (or take a ratio) to see how F_2 relates to F_1.

for question 2)

we know that lamda 1-4 is = lamda2-6 however i figured that
lamda 2 would be equal to lamda one but its not.
You know that the position of the 4th bright line for lambda1 equals the position for the 6th bright line for lambda2. So... what's the formula for the position of the bright lines?
 

DDS

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the formula for question 2 is:

y=m*wavelength*L/d
 

DDS

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As for the first question i get this but i dont know what on earth to do with it:

F= 1/(2L/2) sqrt (3*T)/(U*2)

what i have come up with is

1*srt of 3/2

but i have a feeling thats wrong
 

Doc Al

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DDS said:
the formula for question 2 is:

y=m*wavelength*L/d
Right. Now realize that the only things that change are "m" and wavelength. y, L, & d are fixed.
 

DDS

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so for question 1, can i chose and arbitrary number or jsut make each of them equal to one thus:

4*539=2156

6*539=3234

the ration between them gives me: 1.5
thus the wavelenght changed by 1.5 meaning

(1.5*539) = 808.5 ??
 

DDS

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oops i took the wrong ratio i had a feeling that looked ood. thus the rate of change is 0.66

therefore 0.66*539=359 nm

thanks for the help with question one...how abot question two :biggrin: :tongue2:
 

Doc Al

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DDS said:
As for the first question i get this but i dont know what on earth to do with it:

F= 1/(2L/2) sqrt (3*T)/(U*2)

what i have come up with is

1*srt of 3/2

but i have a feeling thats wrong
Yes, it's wrong. For two reasons:
(1) The mass/length (mu) doesn't merely double; the wire's diameter doubles.
(2) Even assuming that your numbers were OK, your conclusion was not. Given your numbers, you should have concluded:

F_2 = 1/(2L/2) sqrt (3*T)/(U*2) = 2 sqrt(3/2) * 1/(2L) sqrt (T)/(U) = 2 sqrt(3/2) F_1
 

DDS

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so it would be:

F_2 =2 sqrt(3/2) F_1
F_2=2.449*378
f_2=925.9 Hz
 

DDS

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no thats not right...and even if i take the ratio i simply get jsut the answer of

2* srt3/2
 

Doc Al

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DDS said:
no thats not right...and even if i take the ratio i simply get jsut the answer of

2* srt3/2
Of course it's not right. Because it is based on mu doubling, which it doesn't. Notice I prefaced it with "Given your numbers, you should have concluded...". But as I explained, your numbers are not correct.
 

DDS

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so how can u express diameter doubling with the given variables in the question?


as for everyhitng elese, was it correctly done?


meaning it would be 2*sqrt. 3/ something
 

Doc Al

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DDS said:
so how can u express diameter doubling with the given variables in the question?
If the wire diameter doubles what happens to its mass?
 

DDS

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would the mass double as a result?
 

Curious3141

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DDS said:
would the mass double as a result?
Think of the wire as a uniform cylinder. What is the formula for volume of a cylinder in terms of radius and height ? What is the relationship between mass, volume and density ?

Relate the height to the length of the wire segment. Relate the radius to the diameter. Finally, derive an expression for the mass of the wire that's related only to the density, the diameter and the height. You should have a [itex]\pi[/itex] in there. Then, with a little simple algebra, get an expression for [itex]\mu[/itex] in terms of the same things. You should now be able to tie everything up together.
 

DDS

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to me it sounds that you just over complicated a simple problem..

neverthelss i determined that when diameter doubles that u is halved...is this correct??
 

Curious3141

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DDS said:
to me it sounds that you just over complicated a simple problem..
Uhh..not to be unkind, but I'm not the one that can't solve the problem. :rolleyes: I was giving you a basic way to determine what you need, but you can't see that, obviously.

neverthelss i determined that when diameter doubles that u is halved...is this correct??
And this is still not correct.
 

DDS

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Is it 1/4 drop?
 

Doc Al

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Are you suggesting that if you replace the original wire by one twice as thick that the mass per unit length decreases?
 

DDS

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no no no i know its not a drop....i know its a an increase and its either 1/4 or 4x can anyone verify?
 

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