Q 11-14-016NEETJEE MainTop questionMedium
Two successive resonant frequencies of a string fixed at both ends are $420$ Hz and $490$ Hz. Its fundamental frequency is
Answer: (D) $70$ Hz
Concept. For a string fixed at both ends $f_n = nf_1$, so successive resonances differ by $f_1$.
Step 1. Difference. $f_1 = 490 - 420 = 70$ Hz (they are the $6^{\text{th}}$ and $7^{\text{th}}$ harmonics).
Answer. $70$ Hz (Option D).
Solution by Sreeraj P, M.Sc Physics