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

Answer 1

Answer: (A)

For string fixed at both the ends, resonant frequency are given by $f=\frac{mv}{2L}$,
It is given that $315Hz$ and $420Hz$ are two consecutive resonant frequencies, let these are nth and $(n+1)$ th harmonics.
$315=\frac{nv}{2L}$... (i)
$420=\frac{(n+1)v}{2L}$ ... (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{315}{420}=\frac{n}{n+1}$
$\Rightarrow n=3$
From Eq. (i), lowest resonant frequency
$f_0=\frac{v}{2L}$
$=\frac{315}{3}=105Hz$