Consider the situation shown in figure. The wire which has a mass of 5 g oscillates in its second harmonic…
Consider the situation shown in figure. The wire which has a mass of 5 g oscillates in its second harmonic and sets the air column in the tube into vibrations in its fundamental mode. Assuming that the speed of sound in air is 340 ms^-1, find the tension in the wire.
Solution
Sol. Length of wire, L = 50 cm = 0.5 m. Mass of wire = 5 g = 0.005 kg. Linear mass density, $\mu=\dfrac{0.005}{0.5}=0.01\;\mathrm{kg\,m^{-1}}$. For the wire oscillating in its second harmonic, wavelength $\lambda_{wire}=L$ and wave speed on the wire $v_{wire}=\sqrt{T/\mu}$. Frequency of the wire, $f=\dfrac{v_{wire}}{\lambda_{wire}}=\dfrac{\sqrt{T/\mu}}{L}$. The air column vibrates in its fundamental mode (tube closed at one end), so its frequency $f=\dfrac{v_{air}}{4l}$. From the figure the air column length $l=25\;\mathrm{cm}=0.25\;\mathrm{m}$, and $v_{air}=340\;\mathrm{m\,s^{-1}}$. Thus $f=\dfrac{340}{4\times0.25}=340\;\mathrm{Hz}$. Equating frequencies, $\dfrac{\sqrt{T/\mu}}{L}=340$. Substitute $\mu=0.01$ and $L=0.5$: $\dfrac{\sqrt{T/0.01}}{0.5}=340\Rightarrow 20\sqrt{T}=340\Rightarrow\sqrt{T}=17$. Therefore $T=17^2=289\;\mathrm{N}$.
Sol. Frequency of wire, $f_2=\dfrac{2}{2l}\sqrt{\dfrac{T}{\alpha}}=\dfrac{1}{l}\sqrt{\dfrac{T}{M/l}}$
$=\dfrac{1}{0.5}\sqrt{\dfrac{T}{5\times10^{-3}/0.5}}$
$=2\sqrt{100T}=20\sqrt{T}$
Fundamental frequency of closed pipe,
$f_1=\dfrac{v}{4L}=\dfrac{340}{4\times1}=85$
$\therefore\;f_2=f_1\Rightarrow20\sqrt{T}=85$
Therefore, $T=(4.25)^2=18.06\;\mathrm{N}$
Answer: 289 N