Question

Figure shows a tube structure in which a sound signal is sent from one end and is received at the other end. The semicircular part has a radius of $10.0\text{cm}$. The frequency of the sound source can be varied electronically between $1000\text{Hz}$ and $4000\text{Hz}$. Find the approximate frequency(s) at which a maxima of intensity is detected.

[The speed of sound in air = $340{\text{ms}}^{-1}$]

• A. $2110\text{Hz}$
• B. $2980\text{Hz}$
• C. $8940\text{Hz}$
• D. $7840\text{Hz}$

Answer 1

Answer: (B), (C)

$8940\text{Hz}$

The sound wave bifurcates at the junction of the straight and the semicircular parts.

The wave through the straight part travels a distance $l_1=2\times 10=20\text{cm}$ and the wave through the curved part travels a distance $l_2=\pi \times 10=31.4\text{cm}$, before they meet again and travel to the receiver.

The path difference between the two waves received is
$\therefore \Delta l=l_2-l_1=31.4-20=11.4\text{cm}\text{or}0.114\text{m}$

But, for a constructure interference,
Path difference $\left(\Delta l\right)=n\lambda$, where $n=1,2,3,\mathrm{4......}$
$\Rightarrow 0.114=n\times \frac{340}{f}(\because v=f\lambda )$
$\Rightarrow f=n\times \frac{340}{0.114}\approx 2980n\text{Hz}$

For $n=1,f=2980\text{Hz}$
$n=2,f=5960\text{Hz}$
$n=3,f=8940\text{Hz}$

So, from the given options , only options (b) and (c) are correct.