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?
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 = l fA +- fB l
6 = 324 +- fB
fB = 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 ∝ Underroot T
Hence, the beat frequency cannot be 330 Hz
∴ fB = 318 Hz
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
Welcome to the NCERT Solutions for Class 11 Physics - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 18: Two sitar strings A and B playing the note 'Ga' are slightly out of tune and produce beats o....
Comments