Question

A string fixed at both ends has consecutive standing wave modes for which the distances between any two adjacent nodes are $9\text{cm}$ and $8\text{cm}$ respectively. If the tension is $20\text{N}$ and the linear mass density of the string is $8\text{g/m}$, then:

• A. The length of the string is $72\text{cm}$
• B. Fundamental frequency is $35\text{Hz}$
• C. Speed of wave on the string is $100\text{m/s}$
• D. Speed of wave on the string is $50\text{m/s}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A The length of the string is $72\text{cm}$
B Fundamental frequency is $35\text{Hz}$
D Speed of wave on the string is $50\text{m/s}$

Let $l$ be the length of the string and $x$ be the seperation between any two consecutive nodes. Then
For $n$ loops on the string:
$9n=l.........\left(1\right)$
For $n+1$ loops on the string:
$8(n+1)=l.........\left(2\right)$


From $\left(1\right)$ and $\left(2\right)$, we get $n=8$
so, $l=9\times 8=72\text{cm}$
For fundamental frequency,



$\lambda =2l\Rightarrow \lambda =2\times 72=144\text{cm}=1.44\text{m}$

wave speed on the stretched string is given by,
$v=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{20}{8\times {10}^{-3}}}=\sqrt{\frac{{10}^4}{4}}=\frac{100}{2}=50\text{m/s}$
$\therefore$ Fundamental frequency.
$f=\frac{v}{\lambda }=\frac{50}{1.44}=34.72\approx 35\text{Hz}$

Thus, options (a) ,(b) and (d) are the correct answers.