Question

A string of length $3m$ and linear mass density $0.0025kg/m$ is fixed at both ends. One of its resonance frequency is $252Hz$. The next higher resonance frequency is $336Hz$. Then the fundamental frequency will be

• A. $84Hz$
• B. $63Hz$
• C. $126Hz$
• D. $168Hz$

Answer 1

Answer: (A)

$84Hz$

$f_n=n{f}_1=252Hz;f_{n+1}=(n+1)f_1=336Hz$.

Now $\frac{f_n}{f_{n+1}}=\frac{n}{n+1}=\frac{252}{336}\Rightarrow n=3$

$\therefore f_1=\frac{252}{3}=84Hz$