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)

For string fixed at both the ends, resonant frequency are given by $f=\frac{m v}{2 L}$,
It is given that $315 \,Hz$ and $420\, Hz$ are two consecutive resonant frequencies, let these are nth and $(n+1)$ th harmonics.
$315 =\frac{n v}{2 L}$... (i)
$420 =\frac{(n+1) v}{2 L}$ ... (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
$f_{0}=\frac{v}{2 L}$
$=\frac{315}{3}=105\, Hz$