Question

A string fixed at both ends has consecutive standing wave modes for which the distance between adjacent nodes are $12\text{cm}$ and $10\text{cm}$ respectively. If the tension is $20\text{N}$ and the linear mass density is $8\text{g/m}$, what is the fundamental frequency?

• A. $38\text{Hz}$
• B. $41.67\text{Hz}$
• C. $45\text{Hz}$
• D. $50\text{Hz}$

Answer 1

The correct option is B $41.67\text{Hz}$


Let $l$ be the length of the string. Then
$12n=l$ and $10(n+1)=l$
$\Rightarrow n=5$
So, $l=60\text{cm}$
Therefore minimum possible length of string can be $60\text{cm}$.
For the fundamental frequency,
$l=\frac{\lambda }{2}$
$\lambda =2l=2\times 60=120\text{cm}$

Speed of wave on the string,
$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{20}{8\times {10}^{-3}}}=50\text{m/s}$
$\therefore$ fundamental frequency $f$
$f=\frac{v}{\lambda }=\frac{50}{1.2}=41.67\text{Hz}$

Why this question? Tips: In such type of question, the value of $l$ can be assume the minimum possible length of the string.