Physics, asked by ranjeetpratappatil6, 9 months ago

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

Answered by akashdubeypandey
2

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

Answered by Ekaro
18

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

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