Question

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

• A. $10.5\text{Hz}$
• B. $105\text{Hz}$
• C. $155\text{Hz}$
• D. $205\text{Hz}$

Answer 1

The correct option is B $105\text{Hz}$

For a string fixed at both ends, the resonant frequencies are
${\upsilon }_n=\frac{nv}{2L}$ where $n=1,2,3,\dots \dots$

The difference between two consecutive resonant frequencies is
$\Delta {\upsilon }_n={\upsilon }_{n+1}-{\upsilon }_n=\frac{(n+1)v}{2L}-\frac{nv}{2L}=\frac{v}{2L}$
which is also the lowest resonant frequency $(n=1)$.
Thus the lowest resonant frequency for the given string
$=420\text{Hz}-315\text{Hz}=105\text{Hz}$