Two sitar strings A and B playing the note ‘Ga’ are slightly out of tune and producebeats of frequency 6 Hz. The tension in the string A is slightly reduced and thebeat frequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?
It is given that the tuned frequency of the guitar at A is 6 Hz , the original frequency of A is 324 Hz and the reduced beat frequency is 3 Hz .
The train is approaching the platform, so the relative velocity of the sound is the difference in the velocity of sound in air and the speed of the train.
The formula to calculate the new frequency due to tuning of the guitar is,
f 2 = f 1 ± b
Here, the tuned frequency of the guitar is b and the original frequency of A is f 1 .
Substitute the value in the above equation.
f 2 = 324 ± 6 = 330 Hz or 318 Hz
Frequency is directly proportional to the square root of tension. So, on decreasing the tension in the string its highest frequency also changes. As the number of beats are reduced to 3 Hz , the lower frequency remains the same.
The formula to determine the frequency of sound at B is,
f 3 = f 2 + a
Substituting the values in the above equation, we get:
f 3 = 318 + 3 = 321 Hz
Thus, the frequency of B is 321 Hz and the guitar provides 3 beats with the lowest frequency that is 318 Hz .
It is given that the tuned frequency of the guitar at A is
The train is approaching the platform, so the relative velocity of the sound is the difference in the velocity of sound in air and the speed of the train.
The formula to calculate the new frequency due to tuning of the guitar is,
Here, the tuned frequency of the guitar is
Substitute the value in the above equation.
Frequency is directly proportional to the square root of tension. So, on decreasing the tension in the string its highest frequency also changes. As the number of beats are reduced to
The formula to determine the frequency of sound at B is,
Substituting the values in the above equation, we get:
Thus, the frequency of B is
