Question

An organ pipe $A$, with both ends open, has fundamental frequency $300\, Hz$. The third harmonic of another organ pipe $B$, with one end open, has the same frequency as the second harmonic of pipe $A$. The lengths of pipe $A$ and $B$ are (speed of sound in air = $343\, m/s)$

• A. 57.2 cm and 42.9 cm
• B. 57. 2 cm and 45.8 cm
• C. 42.9 cm and 32.2 cm
• D. 42.9 cm and 34.0 cm

Answer 1

Answer: (A)

Given that fundamental frequency of open organ pipe $(A)=300\, H z$
$ f_{1}=300 \,H z=\frac{v}{2 l_{1}} \ldots$ (i)
Frequency of second harmonic-of open organ pipe $(A)=\frac{v}{l_{1}}$
and frequency of third harmonic of closed organ pipe $=\frac{3 v}{4 l_{2}}$
Now according to question $\frac{v}{l_{1}}=\frac{3 v}{4 l_{2}}$
$ \Rightarrow \frac{l_{1}}{l_{2}}=\left(\frac{4}{3}\right) \ldots$ (ii)
Form Eq. (i), we get
$300=\frac{343}{2 l_{1}}$
$l_{1}=\frac{343}{600}=52.16=57.2 \,cm$
Now from Eq. (ii), we get
$l_{2}=\frac{3}{4} \times l_{1}$
$=\frac{3}{4} \times 57.2=42.9 \,cm$