Question

A string is stretched between fixed points 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. Then, the lowest resonant frequency for this string is :

• A. 105 Hz
• B. 1.05 HZ
• C. 1050 HZ
• D. 10.5 HZ

Answer 1

Answer: (A)

105 Hz

Answer 2

For string fixed at both the ends, resonant frequency are given by $f =\frac{n v}{2 L},$where symbols have their usual meanings. It is given that 315 Hz and 420 Hz are two consecutive resonant frequencies, let these are nth and (n + l)th harmonics. $315=\frac{nv}{2L} \, \quad\quad\ldots\left(i\right)$ $420=\frac{\left(n+1\right)v}{2L} \, \quad\ldots$(ii) $=E\left(i\right)+E\left(ii\right) \Rightarrow\frac{315}{420}=\frac{n}{n+1} \Rightarrow n=3$ From E (i), lowest resonant frequency $f _{0}=\frac{v}{2L}=\frac{315}{3}=105 Hz$