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What is the frequency of the vibrator?

  • Thread starter DDS
  • Start date
a.a
127
0
Re: Vibrator

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]
This may be a dumb question, but what do we do after we find n?
 
a.a
127
0
Re: Vibrator

Is it f = nv/2L,
that didnt work for me though.
 

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