# What is the frequency of the vibrator?

1. Jun 12, 2005

### DDS

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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2. Jun 12, 2005

### Pyrrhus

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

$$f = \frac{n}{2L} \sqrt{\frac{T_{1}}{\mu}}$$

for our 16 kg mass

$$f = \frac{n+1}{2L} \sqrt{\frac{T_{2}}{\mu}}$$

3. Jun 12, 2005

### DDS

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

4. Jun 12, 2005

### OlderDan

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?

5. Jun 12, 2005

### Pyrrhus

We get a ratio.

$$\frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}}$$

You can find the tensions.....

6. Jun 12, 2005

### DDS

kind of but not really

7. Jun 12, 2005

### Pyrrhus

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

$$f_{n} = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$

where $$n = 1, 2, 3, ...$$

Last edited: Jun 12, 2005
8. Jun 12, 2005

### DDS

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

9. Jun 12, 2005

### Pyrrhus

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.

10. Jun 13, 2005

### DDS

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

and what do i use for n?

11. Jun 13, 2005

### Pyrrhus

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

12. Jun 13, 2005

### DDS

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

13. Jun 13, 2005

### Pyrrhus

You got

$$\frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}}$$

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

14. Jun 13, 2005

### DDS

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

15. Jun 13, 2005

### Pyrrhus

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.

16. Jun 13, 2005

### DDS

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?

17. Jun 13, 2005

### Pyrrhus

a)

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

$$f = \frac{n}{2L} \sqrt{\frac{T_{1}}{\mu}}$$

for our 16 kg mass

$$f = \frac{n+1}{2L} \sqrt{\frac{T_{2}}{\mu}}$$

We get a ratio.

$$\frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}}$$

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

$$\frac{n+1}{n} = \sqrt{\frac{T_{1}}{T_{2}}}$$

Did you find Tension 1 and Tension 2?, so what $\frac{n+1}{n}$ 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

$$f_{n} = \frac{n}{2L} \sqrt{\frac{T}{\mu}}$$

where $$n = 1, 2, 3, ...$$

Last edited: Jun 13, 2005
18. Jun 13, 2005

### DDS

for part A i get :

348.7 Hz

B)

397.4 kg

what did you get?

19. Jun 13, 2005

### OlderDan

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.

20. Jun 13, 2005

### DDS

part i get 349.3Hz

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