Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
Solution and Explanation
Frequency of string A, fA = 324 Hz
Frequency of string B = fB
Beat’s frequency, n = 6 Hz
beat's frequency is given as :
\(n=|f_A±f_B|\)
\(6=324±f_B\)
\(f_B=330\, Hz \,or \,318\, Hz\)
Frequency decreases with a decrease in the tension in a string. This is because frequency is directly proportional to the square root of tension. It is given as:
\(v∝√T\)
Hence the beat frequency cannot be 330 Hz
\(∴f_B=318 Hz\)
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Concepts Used:
Beats
What is Beat Frequency?
Let’s see from fig.m, the frequency of a pink-colored wave is f1, and that of a green-colored wave is f2. So, the frequency of the beat is the difference between these two, which is:
fBEATS = |f1 - f2|
Interference and Beats
When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.
What is Interference?
When two or more waves travelling in a medium meet, the resulting phenomenon is called interference and beats are an excellent example of the phenomenon of interference.
The Application of Beats:
- Beats are used in determining the unknown frequency
- Beats are used in determining the existence of poisonous gases in mines.