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In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6Hz. When tension in B is slightly decreased, the beat frequency increases to 7Hz. If the frequency of A is 530Hz, the original frequency of B will be
- a)524Hz
- b)536Hz
- c)537Hz
- d)523Hz
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
In a guitar, two strings A and B made of same material are slightly o...
Solution:
Given, strings A and B produce beats of frequency 6Hz when out of tune. Let the frequency of string B be 'f'.
When the tension in string B is decreased, the beat frequency increases to 7Hz. Let the new frequency of string B be 'f'.
The beat frequency is given by the difference in frequencies of the two strings:
7 Hz = |530 Hz - f|
As the beat frequency is positive, we can write:
f = 530 Hz - 7 Hz = 523 Hz
Therefore, the original frequency of string B was 523 Hz.
When the tension in string B is decreased, the beat frequency increases from 6Hz to 7Hz.
The change in beat frequency is given by:
Δf = |f1 - f2| = |6 Hz - 7 Hz| = 1 Hz
As the beat frequency increases when the tension in string B is decreased, we can conclude that the frequency of string B was higher than the original frequency of 523 Hz.
Let the new frequency of string B be 'f'.
When the tension in string B is decreased, the frequency of string B decreases.
Therefore, we can write:
f = 523 Hz - Δf = 523 Hz - 1 Hz = 522 Hz
Therefore, the new frequency of string B is 522 Hz.
We can verify the result by calculating the beat frequency between strings A and B:
Beat frequency = |530 Hz - 522 Hz| = 8 Hz
As the beat frequency is not equal to 6 Hz, we can conclude that the result is incorrect.
From the given information, we know that the beat frequency between strings A and B is 6 Hz.
Therefore, we can write:
6 Hz = |530 Hz - f|
As the beat frequency is positive, we can write:
f = 530 Hz - 6 Hz = 524 Hz
Therefore, the original frequency of string B was 524 Hz (Option A).
Given, strings A and B produce beats of frequency 6Hz when out of tune. Let the frequency of string B be 'f'.
Step 1: Finding the Original Frequency of B
When the tension in string B is decreased, the beat frequency increases to 7Hz. Let the new frequency of string B be 'f'.
The beat frequency is given by the difference in frequencies of the two strings:
7 Hz = |530 Hz - f|
As the beat frequency is positive, we can write:
f = 530 Hz - 7 Hz = 523 Hz
Therefore, the original frequency of string B was 523 Hz.
Step 2: Finding the Change in Frequency of B
When the tension in string B is decreased, the beat frequency increases from 6Hz to 7Hz.
The change in beat frequency is given by:
Δf = |f1 - f2| = |6 Hz - 7 Hz| = 1 Hz
As the beat frequency increases when the tension in string B is decreased, we can conclude that the frequency of string B was higher than the original frequency of 523 Hz.
Step 3: Finding the New Frequency of B
Let the new frequency of string B be 'f'.
When the tension in string B is decreased, the frequency of string B decreases.
Therefore, we can write:
f = 523 Hz - Δf = 523 Hz - 1 Hz = 522 Hz
Therefore, the new frequency of string B is 522 Hz.
Step 4: Verifying the Result
We can verify the result by calculating the beat frequency between strings A and B:
Beat frequency = |530 Hz - 522 Hz| = 8 Hz
As the beat frequency is not equal to 6 Hz, we can conclude that the result is incorrect.
Step 5: Correcting the Result
From the given information, we know that the beat frequency between strings A and B is 6 Hz.
Therefore, we can write:
6 Hz = |530 Hz - f|
As the beat frequency is positive, we can write:
f = 530 Hz - 6 Hz = 524 Hz
Therefore, the original frequency of string B was 524 Hz (Option A).
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Community Answer
In a guitar, two strings A and B made of same material are slightly o...
Frequency of string,
Frequency ∝ √ Tension
Difference of fA and fB is 6 Hz.
If tension decreases, fB decreases and becomes fB′
Now, difference of fA and fB′ = 7Hz (increases) So,fA > fB
fA − fB = 6Hz
⇒ fA = 530Hz
⇒fB = 524Hz (original)
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In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6Hz. When tension in B is slightly decreased, the beat frequency increases to 7Hz. If the frequency of A is 530Hz, the original frequency of B will bea)524Hzb)536Hzc)537Hzd)523HzCorrect answer is option 'A'. Can you explain this answer?
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In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6Hz. When tension in B is slightly decreased, the beat frequency increases to 7Hz. If the frequency of A is 530Hz, the original frequency of B will bea)524Hzb)536Hzc)537Hzd)523HzCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6Hz. When tension in B is slightly decreased, the beat frequency increases to 7Hz. If the frequency of A is 530Hz, the original frequency of B will bea)524Hzb)536Hzc)537Hzd)523HzCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency 6Hz. When tension in B is slightly decreased, the beat frequency increases to 7Hz. If the frequency of A is 530Hz, the original frequency of B will bea)524Hzb)536Hzc)537Hzd)523HzCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
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