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wel
3 years ago
14

A string of length L, mass per unit length \mu, and tension T is vibrating at its fundamental frequency. What effect will the fo

llowing have on the fundamental frequency?
(a) The length of the string is doubled, with all other factors held constant ?

(b) The mass per unit length is doubled, with all other factors held constant.

(c) The tension is doubled, with all other factors held constant.
Physics
1 answer:
viva [34]3 years ago
8 0

The fundamental frequency on a vibrating string is given by:

f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}

where

L is the length of the string

T is the tension

\mu is the mass per unit length of the string

Keeping this equation in mind, we can now answer the various parts of the question:

(a) The fundamental frequency will halve

In this case, the length of the string is doubled:

L' = 2L

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2(2L)}\sqrt{\frac{T}{\mu}}=\frac{1}{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{2}

So, the fundamental frequency will halve.

(b) the fundamental frequency will decrease by a factor \sqrt{2}

In this case, the mass per unit length is doubled:

\mu'=2\mu

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{T}{2 \mu}}=\frac{1}{\sqrt{2}}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\frac{f}{\sqrt{2}}

So, the fundamental frequency will decrease by a factor \sqrt{2}.

(c) the fundamental frequency will increase by a factor \sqrt{2}

In this case, the tension is doubled:

T'=2T

Substituting into the expression of the fundamental frequency, we find the new frequency:

f'=\frac{1}{2L}\sqrt{\frac{2T}{\mu}}=\sqrt{2}(\frac{1}{2L}\sqrt{\frac{T}{\mu}})=\sqrt{2}f

So, the fundamental frequency will increase by a factor \sqrt{2}.

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