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Simple wave question

"the fundamental frequency of vibration of a particular string is

f

. What would the fundamental frequency be if the length of the wire was halved and the tension were to be increased by a factor of 4?"

thanks :smile:

I know we use

f=\frac{n}{2L}\times \frac{T}{\mu}

(edited 11 years ago)

Reply 1

Linked

Original post by upthegunners

"the fundamental frequency of vibration of a particular string is

f

. What would the fundamental frequency be if the length of the wire was halved and the tension were to be increased by a factor of 4?"

thanks :smile:

I know we use

f=\frac{n}{2L}\times \frac{T}{\mu}

4 to the power of 2 = 16 times greater.

Reply 2

upthegunners

Original post by Linked

4 to the power of 2 = 16 times greater.

why is it

4^2

Reply 3

dasdaa

Original post by upthegunners

why is it

4^2

Im not really sure. Was the answer right?

Reply 4

dasdaa

Original post by dasdaa

Im not really sure. Was the answer right?

Im not sure why he said that? Its wrong!

Reply 5

Stonebridge

Original post by upthegunners

"the fundamental frequency of vibration of a particular string is

f

. What would the fundamental frequency be if the length of the wire was halved and the tension were to be increased by a factor of 4?"

thanks :smile:

I know we use

f=\frac{n}{2L}\times \frac{T}{\mu}

Check the formula you have used.
It isn't correct.

Reply 6

upthegunners

Original post by Stonebridge

Check the formula you have used.
It isn't correct.

Hi Stonebridge

Sorry I have just realised I typed a latex code in wrong.

There should be a sqrt sign over the last fraction in the equation.

I am just not sure how to work out the frequency as we have been given no numbers.

Reply 7

Stonebridge

Original post by upthegunners

The (correct) formula says that the frequency depends on the square root of the tension.
It also says it depends on 1/length of wire.

Take tension first.
If it depends on the square root of tension, then increasing the the tension by any factor will increase the frequency by the square root of that factor.
So 25 times the tension would give you 5 times the frequency, for example.

Taking length next, inverse relationship (1/L) means if you double L you half the frequency, for example. If you were to have 1/3 the length you would have 3 times the frequency.

Can you apply this to your problem now? You have to combine both changes. Tension and length, of course.