Question

A string is stretched between fixed point separated by 75.0 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is :

• A. 105 Hz
• B. 155 Hz
• C. 205 Hz
• D. 10.5 Hz

Answer 1

The correct option is A 105 Hz

here, $315=\frac{n\nu }{2L}$ and $420=\frac{(n+1)\nu }{2L}$

so, $\frac{420}{315}=\frac{n+1}{n}\Rightarrow n=3$

Lowest resonant frequency will be $\frac{1}{3}$ of $315=\frac{315}{3}=105Hz$