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)

$105Hz$

Given $\frac{nv}{2l}=315$ and $(n+1)\frac{v}{2l}=420$
$\Rightarrow \frac{n+1}{n}=\frac{420}{315}\Rightarrow n=3$
Hence $3\times \frac{v}{2l}=315\Rightarrow \frac{v}{2l}=105Hz$
The lowest resonant frequency is when $n=1$
Therefore lowest resonant frequency $105Hz$.