Question

If the length of an open organ pipe is $33.3cm$, then the frequency of fifth overtone is $(v_{sound}=333m{s}^{-1})$

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

Answer 1

Answer: (A)

An open organ pipe is open at both ends.
Wavelength of vibration in open organ pipe is given by
$\lambda =\frac{2L}{n},$ where $n=1,2,3$, ...
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 or sound of this frequency is called fundamental note or first harmonic.
Similarly for
$n=2,v_2=\frac{2v}{2L}=2v_1$
For $n=3,v_3=\frac{3v}{2L}=3v_1$
In general, the frequency of vibration in nth normal mode of vibration in open organ pipe would be
$v_n=n{v}_1$
Hence, frequency of fifth overtone
$=6\times$ fundamental frequency
$=6\times \frac{v}{2L}$
$=\frac{6\times 333}{2\times 33.3\times 10^{-2}}=3000Hz$