Q 11-14-004NEETNEET 2020Top questionMedium
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 :
Answer: (B) $524$ Hz
With a beat frequency of $6$ Hz, $f_B = 530 \pm 6 = 524$ Hz or $536$ Hz.
Decreasing the tension in B lowers its frequency ($f \propto \sqrt{T}$).
If $f_B = 536$ Hz, lowering it would bring it closer to $530$ Hz and the beats would decrease. If $f_B = 524$ Hz, lowering it moves it further from $530$ Hz, and the beats increase to $7$ Hz. ✓
So $f_B = 524$ Hz.
Solution by Sreeraj P, M.Sc Physics