Question

If the length of an open organ pipe is $33.3cm$, then the frequency of fifth overtone is: $(v_{sound}=333m/s)$

• A. $3000Hz$
• B. $1500Hz$
• C. $2500Hz$
• D. $1250Hz$

Answer 1

The correct option is A $3000Hz$

In an open organ pipe, all harmonics are present.
An open organ pipe is open at both ends.
Wavelength of vibration in open organ pipe is given by $\lambda =\frac{2L}{n}$, where $n=1,2,3,\dots$
Let ${\lambda }_1$ be the wavelength of stationary waves set up in the open organ pipe corresponding to $n=1$.
${\lambda }_1=\frac{2L}{1}$ or $L=\frac{{\lambda }_1}{2}$
The frequency of vibration in this mode is given by
$v_1=\frac{v}{{\lambda }_1}=\frac{v}{2L}$
or, $v_1=\frac{v}{2L}$
This is the lowest frequency of vibration and is called the fundamental frequency. The note of sound of this frequency is called fundamental note or first harmonic.
Similarly for
$n=2$, $v_2=\frac{2v}{2L}=2v_1$
for $n=3$, $v_3=\frac{3v}{2L}=3v_2$
In general, the frequency of vibration in $n$th normal mode of vibration in open organ pipe would be
$v_n=n{v}_1$
Hence, frequency of fifth overtone
$=6\times$ fundamental frequency
$=6\times \frac{v}{2L}$
$=\frac{6\times 333}{2\times 33.3\times {10}^{-2}}$
$=3000Hz$