Question

The length of a sonometer wire is $1.21\text{m}$. It is divided into three segments. What should be the lengths of the three segments for their fundamental frequencies to be in the ratio $1:2:3$?

• A. $0.66\text{m},0.33\text{m},0.22\text{m}$
• B. $0.72\text{m},0.30\text{m},0.18\text{m}$
• C. $0.56\text{m},0.44\text{m},0.21\text{m}$
• D. $0.82\text{m},0.18\text{m},0.21\text{m}$

Answer 1

The correct option is A $0.66\text{m},0.33\text{m},0.22\text{m}$

Given that,
Length of the sonometer wire, $L=1.21\text{m}$
$\Rightarrow f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$


Suppose the lengths of the three segments are $L_1,L_2$ and $L_3$ respectively. The fundamental frequencies are,
$f_1=\frac{1}{2L_1}\sqrt{\frac{T}{\mu }}...\left(A\right)$
$f_2=\frac{1}{2L_2}\sqrt{\frac{T}{\mu }}...\left(B\right)$
$f_3=\frac{1}{2L_3}\sqrt{\frac{T}{\mu }}...\left(C\right)$

Given, $f_1:f_2:f_3=1:2:3...\left(i\right)$
and $f_1{L}_1=f_2{L}_2=f_3{L}_3..\left(ii\right)$
[From $A,B$ and $C$]

From the equation $\left(i\right)$, we can write.
$f_2=2f_1$ and $f_3=3f_1$
So, by equation $\left(ii\right)$,
$L_2=\frac{f_1}{f_2}{L}_1=\frac{L_1}{2}$
& $L_3=\frac{f_1}{f_3}{L}_1=\frac{L_1}{3}$

$L_1+L_2+L_3=1.21\text{m}$
$\Rightarrow L_1+\frac{L_1}{2}+\frac{L_1}{3}=1.21\text{m}\Rightarrow L_1=0.66\text{m}$
$\therefore L_2=\frac{L_1}{2}=\frac{0.66}{2}=0.33\text{m}$
& $L_3=\frac{L_1}{3}=\frac{0.66}{3}=0.22\text{m}$
Thus, option (a) is correct.