Question

Two strings of a guitar are of lengths 40 cm and 20 cm, such that they are made from same material and are of same thickness and under same tension. If the frequency of longer wire is 250 Hz. Find the frequency of smaller wire.

Answer 1

For frequency in a closed organ plane,

$f = \frac{nv}{2l}$


Using this relation for 40 cm i.e 040 m string,
$f = \frac{nv}{0.80} = 250 {s}^{ - 1} \\ \\ nv = 200 {s}^{ - 1}$


For 20 cm i.e 0.20 m string,

$\frac{nv}{0.4} \\ \\ = \frac{200}{0.4} \\ \\ = 500 {s}^{ - 1}$


Hence your answer.

Answer 2

The frequency of smaller wire is 5000 Hz.

Explanation:

Given that,

Length of longer wire = 40 cm

Length of smaller wire = 20 cm

Frequency of longer wire = 250 Hz

We know that,

The frequency of wire is

$f =\dfrac{nv}{2l}$

We need to calculate the frequency of smaller wire

Using formula of frequency for two string

$\dfrac{f_{2}}{f_{1}}=\dfrac{\dfrac{nv}{2l_{2}}}{\dfrac{nv}{2l_{1}}}$

$\dfrac{f_{2}}{f_{1}}=\dfrac{l_{1}}{l_{2}}$

Where, $f_{1}$ = frequency of longer wire

$f_{2}$ = frequency of smaller wire

$l_{1}$ = length of longer wire

$l_{2}$ = length of smaller wire

Put the value into the formula

$\dfrac{f_{2}}{250}=\dfrac{40}{20}$

$f_{2}=20\times250$

$f_{2}=5000\ Hz$

Hence, The frequency of smaller wire is 5000 Hz.

Learn more :

Topic = frequency

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