Question

the linear density of stretched string is increased by 1%.the change in fundamental frequency of string is
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Answer 1

Answer:

Given:

Linear density of stretched string is increased by 1%.

To find:

Change in frequency of string.

Concept:

Due to change in linear density of a string , it leads to change in the velocity of the transverse pulse (wave) .

Now changes in the Velocity of the wave leads to changes in the fundamental frequency of the standing wave.

Calculation:

Let linear density be μ.

Velocity of wave is given as :

$v = \sqrt{\dfrac{T}{ \mu}}$

So fundamental frequency will be :

$frequency = \dfrac{velocity}{wavelength}$

$= > freq. = \dfrac{ \sqrt{\frac{T}{ \mu}} }{ \lambda}$

So for small changes in μ , we can say that :

$\boxed{ \large{ \dfrac{\Delta f}{f} = \dfrac{1}{2} \dfrac{\Delta \mu}{ \mu} }}$

$\large{ = > \dfrac{ \Delta f}{f} = \dfrac{1}{2} \times 1\% }$

$\large{ = > \dfrac{\Delta f}{f} = 0.5\% }$

So final answer :

$\boxed{ \huge{ \red{ \bold{ \dfrac{\Delta f}{f} = 0.5\% }}}}$

Answer 2

$</p><p>\huge{\tt{\pink{Hello!!! }}}$

$\tt{ \red{let \: the \: linear \: density \: be \: \nu \: }}$

We Know That :

$\tt{ \blue{ \boxed{ \boxed{frequency \: = \: \frac{velocity}{wavelength} \: = \: \frac{v}{ \lambda} }}}}$

$\tt{ \purple{ \boxed{ \boxed{v \: = \: \sqrt{ \frac{t}{ \nu} } }}}}$

$\tt{ \green{ \boxed{ \boxed{frequency \: = \: \frac{ \sqrt{ \frac{t}{ \nu} } }{ \lambda} }}}}$

For Small Values :

$\tt{ \purple{ \frac{ \delta \: f}{f} = \frac{1}{2} \times \frac{ \delta \: \nu}{nu} = \: \frac{1}{2} \times 1 \% \: = \: 0.5\% \: }}$

So Change In Fundamental Frequency Of String Is 0.5%