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. The lowest resonant frequency for this string is

• A. $105Hz$
• B. $155Hz$
• C. $205Hz$
• D. $10.5Hz$

Answer 1

The correct option is A $105Hz$

For string fixed at both ends,
$f=\frac{nv}{2L}$

Length of the string $L=75cm$

Since the given resonance frequencies are consecutive, we can write

$\frac{xv}{2L}=315$
$\frac{(x+1)v}{2L}=420$
for some integer $x>1$

Fundamental frequency $f_0=\frac{v}{2L}=\frac{(x+1)v}{2L}-\frac{xv}{2L}=420-315=105Hz$