Published by:
CGP EDU Academic Team
Published on: August 21, 2026
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:
Text Solution
Verified by ExpertsThe correct answer is:
Given that,
Difference of f A and f B ,
f = 6 Hz
If tension decreases,
decreases and becomes f’b.
Now, difference of f A and f B ,
f '= 7 Hz
f A - f B = 6 Hz
Now, since f A = 530 Hz
f B = 524 Hz
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale.
The pitc…
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over …
An electron is accelerated from rest through a potential difference of V volt If the de Broglie wav…
In a certain region of space with volume 0.2 m 3 , the electric potential is found to be 5 V throug…
A cylinder contains hydrogen gas at pressure of 249 kPa and temperature 27°C.
Its density is : (R =…
The mean free path for a gas, with molecular diameter d and number density n can be expressed as :