In a guitar, two strings A and B made of same
material are slightly out of tune and produce beats
of frequency 6 Hz. When tension in B is slightly
decreased, the beat frequency increases to 7 Hz.
If the frequency of A is 530 Hz, the original
frequency of B will be:
Answers
Explanation:
Frequency of string A, f
A
=324Hz
Frequency of string B=f
B
Beat’s frequency, n=6Hz
Beat's Frequency is given as:
n=∣f
A
−f
B
∣
6=∣324−f
B
∣
f
B
=330Hz or 318Hz
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:
f∝
T
Hence, the beat frequency cannot be 330 Hz
∴f
B
=318Hz
Answer :
Frequency of string A (f₁) = 530Hz
Beat’s frequency (n) = 6 Hz
When tension in B is slightly decreased, the beat frequency increases to 7 Hz.
We have to find the frequency of string B
_________________________________
◈ Beat's Frequency is given by
- n = | f₁ - f₂ |
Let frequency of string B be f₂.
⇒ 6 = 530 ± f₂
⇒ f₂ = 524 Hz or 536 Hz
We know that,
Frequency decreases with a decrease in the tension in a string because frequency is directly proportional to the square root of tension.
- f ∝ √T
Hence, the beat frequency cannot be 536 Hz.
Frequency of string B (f₂) = 524 Hz