Question

A tube $1.20\text{m}$ long is closed at one end. A stretched wire is placed near the open end. The wire is $0.330\text{m}$ long and has a mass of $9.60\text{g}$. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at the fundamental frequency. Then:
[Take speed of sound in air as $343\text{m/s}$]

• A. Fundamental frequency of air column is $71.45\text{Hz}$.
• B. Fundamental frequency of air column is $35\text{Hz}$.
• C. The tension in the wire is $29.5\text{N}$.
• D. The tension in the wire is $64.50\text{N}$.

Answer 1

Answer: (A), (D)

The tension in the wire is $64.50\text{N}$.

Given that,
Length of tube $\left(L\right)=1.20\text{m}$
Length of wire $\left(l\right)=0.330\text{m}$
Mass of wire $\left(m\right)=9.60\text{g}$
Speed of sound $\left(v_s\right)=343\text{m/s}$

Frequency of closed organ pipe is given by
$f_n=\frac{(2n-1)v}{4l}\text{where}n=1,2,3,\mathrm{4....}$
Fundamental frequency, $f=\frac{v}{4L}\Rightarrow f=\frac{343}{4\times 1.20}=71.45\text{Hz}$

Now, linear mass density of wire,
$\mu =\frac{m}{l}=\frac{9.60\times {10}^{-3}}{0.33}=0.029\text{kg/m}$

Frequency of wave on the wire is given by $f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$
Fundamental frequency of wave on the wire
$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$
$\Rightarrow T={(f\times 2l)}^2\times \mu$
Substituting the given data, we get
$T={(71.45\times 2\times 0.33)}^2\times 0.029$
[since frequency of oscillation of wire is the same as the fundamental frequency of air column]
$\Rightarrow T=64.48\simeq 64.50\text{N}$
Hence, options (a) and (d) are correct.