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3 years ago

My next photo How's it

Physics

1 answer:

allochka39001 [22]3 years ago

3 0

Answer:

Pretty good

Explanation:

is there nxt 1

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

viva [34]

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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