
Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats of frequency $6Hz$. Tension in the string A is slightly reduced and the beat frequency is found to reduce to $3Hz$. If the original frequency of A is $324Hz$ what is the frequency of $B$?
Answer
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Hint: The above problem is based on the beat frequency concept, which is given by the formula:
$n = \left| {{F_A} - {F_B}} \right|$ ($F_A$ frequency of string A and $F_B$ frequency of string B and n is the beat frequency)
Beat is the alternate waning and waxing of sound produced by strings of musical instruments.
Using the above relation we will calculate the frequency of string $B$.
Complete step by step solution:
First we will discuss beats and beat frequency.
When two sounding bodies of nearly the same frequency and same amplitude are sounded together, the resultant sound comprises alternate maxima and minima.
The phenomenon of alternate waxing and waning of sound at regular intervals is called Beat and the number of beats heard per second is called beat frequency. It is equal to the difference in frequency between sounding bodies. Beats are heard only when the difference in frequencies of two sounding bodies is not more than ten; this is due to persistence of sound.
Now we will calculate the frequency of string (B).
Frequency of string $A$ = $324Hz$
Beat Frequency = $6Hz$
$n = \left| {{F_A} - {F_B}} \right|$......................(1) (two values of beat frequency will come out from the modulus function positive and negative)
On substituting the values of frequencies in equation 1
$\Rightarrow 6 = 324 - {F_B}$
$\Rightarrow 6 = - 324 + {F_B} $
$ \Rightarrow {F_B} = 318Hz, 330Hz$ (we have addition and subtraction)
In the question we are given that the tension of the string decreases which means that frequency also decreases. Therefore, frequency which is less in value will be considered.
$F_B$ = $318 Hz$
Note: Various applications of beats: To determine the unknown frequency of tuning fork by comparing with the known frequency, use in music( used for tuning musical instruments), Used in electronics beat frequency oscillators are used, Used in mines; to determine the presence of dangerous gases in mines.
$n = \left| {{F_A} - {F_B}} \right|$ ($F_A$ frequency of string A and $F_B$ frequency of string B and n is the beat frequency)
Beat is the alternate waning and waxing of sound produced by strings of musical instruments.
Using the above relation we will calculate the frequency of string $B$.
Complete step by step solution:
First we will discuss beats and beat frequency.
When two sounding bodies of nearly the same frequency and same amplitude are sounded together, the resultant sound comprises alternate maxima and minima.
The phenomenon of alternate waxing and waning of sound at regular intervals is called Beat and the number of beats heard per second is called beat frequency. It is equal to the difference in frequency between sounding bodies. Beats are heard only when the difference in frequencies of two sounding bodies is not more than ten; this is due to persistence of sound.
Now we will calculate the frequency of string (B).
Frequency of string $A$ = $324Hz$
Beat Frequency = $6Hz$
$n = \left| {{F_A} - {F_B}} \right|$......................(1) (two values of beat frequency will come out from the modulus function positive and negative)
On substituting the values of frequencies in equation 1
$\Rightarrow 6 = 324 - {F_B}$
$\Rightarrow 6 = - 324 + {F_B} $
$ \Rightarrow {F_B} = 318Hz, 330Hz$ (we have addition and subtraction)
In the question we are given that the tension of the string decreases which means that frequency also decreases. Therefore, frequency which is less in value will be considered.
$F_B$ = $318 Hz$
Note: Various applications of beats: To determine the unknown frequency of tuning fork by comparing with the known frequency, use in music( used for tuning musical instruments), Used in electronics beat frequency oscillators are used, Used in mines; to determine the presence of dangerous gases in mines.
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