Question

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

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

Answer 1

Answer: (A)

For string fixed at both the ends, resonant frequency are given by $f=\frac{nv}{2L},$ 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.
$315=\frac{nv}{2L}...\left(i\right)$
$420=\frac{\left(n+1\right)v}{2L}...\left(ii\right)$
$\Rightarrow Eq.\left(i\right)÷Eq.\left(ii\right)\Rightarrow \frac{315}{420}=\frac{n}{n+1}\Rightarrow n=3$
From $Eq.\left(i\right),$ lowest resonant frequency
$f_0=\frac{v}{2L}=\frac{315}{3}=105Hz$