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What is the frequency of the vibrator?

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DDS

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In the arrangement shown in the figure, a mass can be hung from a string (with a linear mass density of μ=0.00165 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f), and the length of the string between point P and the pulley is L=2.00 m. When the mass m is either 15.9 kg or 24.8 kg, standing waves are observed; however no standing waves are observed with any mass between these values.

What is the frequency of the vibrator? (Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.)

B)What is the largest mass for which standing waves could be observed?

For part A i have found the velocity and i am thinking of equationg two equations to find the wavelenghts of both masses and then finding the frequencies of of those masses.

But i do not know how to solve the problem mathematically


As for part B, i am completely lost on how to determine the mass.


Please can anyone give me some detialed help
 

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Pyrrhus

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I will start your off.

Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.


also

constant frequency f

If n is the number of nodes for the wave produced with the mass 25 kg, then n+1 is the number of nodes for the wave produced with the mass 16kg,

so, for our 25kg mass

[tex] f = \frac{n}{2L} \sqrt{\frac{T_{1}}{\mu}} [/tex]

for our 16 kg mass

[tex] f = \frac{n+1}{2L} \sqrt{\frac{T_{2}}{\mu}} [/tex]
 

DDS

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i got exactly to what you showed me but i dont know where to take it from there
 

OlderDan

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DDS said:
In the arrangement shown in the figure, a mass can be hung from a string (with a linear mass density of μ=0.00165 kg/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f), and the length of the string between point P and the pulley is L=2.00 m. When the mass m is either 15.9 kg or 24.8 kg, standing waves are observed; however no standing waves are observed with any mass between these values.

What is the frequency of the vibrator? (Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.)

B)What is the largest mass for which standing waves could be observed?

For part A i have found the velocity and i am thinking of equationg two equations to find the wavelenghts of both masses and then finding the frequencies of of those masses.

But i do not know how to solve the problem mathematically


As for part B, i am completely lost on how to determine the mass.


Please can anyone give me some detialed help
In part a) you say you have calculated the velocity. If so you must have two velocities, one for each mass. All you know is that both ends of the 2 meter length are nodes. You do not know how many are in between, but you should know what the difference is between the number of nodes for the first mass and the number for the second mass. A solution to a) will tell you how many there are in each case. Once you know that you can figure out the wavelengths involved. You should know the wavelength when each end is a node with an antinode in the middle, which is the case for part b). Do you understand why?
 

Pyrrhus

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We get a ratio.

[tex] \frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}} [/tex]

You can find the tensions.....
 

DDS

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kind of but not really
 

Pyrrhus

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For B is quite intuitive,

Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.

The greatest tension will have only 1 loop (1 normal mode), you know why?

Consider the relation

[tex] f_{n} = \frac{n}{2L} \sqrt{\frac{T}{\mu}} [/tex]

where [tex] n = 1, 2, 3, ... [/tex]
 
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DDS

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when i go to find the tension what do i assume n is and am i finding the ratio of tension because i dont know either value T1 or T2
 

Pyrrhus

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The tension is a massles string is the same throughout so, if the weight produced by each mass (25 and 16) will give each tension.
 

DDS

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so when i solve that equation i get the tension that exists in both cases?

and what do i use for n?
 

Pyrrhus

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Why don't you keep working that relation above, and see where it takes you?... :rolleyes:
 

DDS

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the thing is i dont know where to take it from there
 

Pyrrhus

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You got

[tex] \frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}} [/tex]

Did you find Tension 1 and Tension 2?, so what [itex] \frac{n+1}{n} [/itex] equals to? can you now find n?
 

DDS

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i thaught teh n expression will help me find tension...now im really confused...

can we go back to waht i had and what u showed me to start me off...where do i go from there
 

Pyrrhus

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DDS, i practically handed you over answers for a) and b), you need to go back and read all again, i know you can understand what i did. If i continue i will just write the answers.
 

DDS

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no i dont want the answers that doesnt due me any good, its just im getting confused. Maybe its because ive tried t oapprocahc this problem in so many ways today its just when i progress all my previous tactics come back and screw me.

is there anything you can tell me that wont give the answer?
 

Pyrrhus

Homework Helper
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Here, read this

a)
I will start your off.

Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.


also

constant frequency f

If n is the number of nodes for the wave produced with the mass 25 kg, then n+1 is the number of nodes for the wave produced with the mass 16kg,

so, for our 25kg mass

[tex] f = \frac{n}{2L} \sqrt{\frac{T_{1}}{\mu}} [/tex]

for our 16 kg mass

[tex] f = \frac{n+1}{2L} \sqrt{\frac{T_{2}}{\mu}} [/tex]

We get a ratio.

[tex] \frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}} [/tex]

You can find the tensions.....

The tension is a massles string is the same throughout so, if the weight produced by each mass (25 and 16) will give each tension.

You got

[tex] \frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}} [/tex]

Did you find Tension 1 and Tension 2?, so what [itex] \frac{n+1}{n} [/itex] equals to? can you now find n?

b)

For B is quite intuitive,

Hint: The greater the tension in the string the smaller the number of nodes in the standing wave.

The greatest tension will have only 1 loop (1 normal mode), you know why?

Consider the relation

[tex] f_{n} = \frac{n}{2L} \sqrt{\frac{T}{\mu}} [/tex]

where [tex] n = 1, 2, 3, ... [/tex]
 
Last edited:

DDS

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for part A i get :

348.7 Hz


B)

397.4 kg

what did you get?
 

OlderDan

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DDS said:
for part A i get :

348.7 Hz


B)

397.4 kg

what did you get?
Part A does not look right to me. Part B looks very close, so I assume we only differ by a round off. Check part A, and if you still think yours is right tell me what you got for n and n + 1.
 

DDS

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part i get 349.3Hz

and B i guess it could be 397 or is 397.4 the answer?
 

OlderDan

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DDS said:
part i get 349.3Hz

and B i guess it could be 397 or is 397.4 the answer?
What did you get for n and n + 1?
 

DDS

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part b i corrected and got 398.2 kg

but why am i getting part a wrong
 

OlderDan

Science Advisor
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DDS said:
part b i corrected and got 398.2 kg

but why am i getting part a wrong
How am I suppose to know that? You don't show how you got the answer and you refuse to answer my questions. Good night.
 
How did you do part b without knowing the correct answer for part A)? As OldDan says, can you please show us your work so we can explain to you where you went wrong?
 

DDS

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I figured it out thanks everyone
 

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