# What is the frequency of the vibrator?

a.a
Re: Vibrator

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 $$n = 1, 2, 3, ...$$
This may be a dumb question, but what do we do after we find n?

a.a
Re: Vibrator

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