Question

An organ pipe closed at one end has fundamental frequency of$1500Hz$. The maximum number of overtones generated by this pipe which a normal person can hear is:
(A) $4$
(B) $13$
(C) $6$
(D) $9$

Answer 1

Hint We are given with a one end closed organ pipe of a fundamental frequency and are asked to find the maximum number of overtone generated by this pipe which a normal person can hear or in other words, we will have to find the number of overtones lying in the audible frequency range of the a normal person. Thus, we will formulate the equation of the frequency of all the harmonics.

Complete step by step answer
For a one end open organ pipe, the closed end corresponds to one node of the wave and the open end corresponds to an antinode of the wave.
Thus,
The length will correspond to multiples of one-fourth of the wavelength of the wave.
Thus,
$L = (2n + 1)\dfrac{\lambda }{4}$
Thus,
$\lambda = \dfrac{{4L}}{{(2n + 1)}}$
Now,
We know,
$v = \lambda f$
Where,$v$ is the speed of the wave,$\lambda$ is the wavelength of the wave and$f$ is the frequency of the frequency.
Further,
$f = \dfrac{v}{\lambda }$
Substituting the value of wavelength, we get
${f_n} = (2n + 1)\dfrac{v}{{4L}}$
Now,
The fundamental frequency of the situation is when$n = 0$,
Thus,
${f_0} = \dfrac{v}{{4L}}$
By question,
${f_0} = 1500Hz$
Thus,
Substituting this value, we get
$1500 = \dfrac{v}{{4L}}$
Thus,
$\dfrac{v}{L} = 6000$
Now,
For first overtone,
$n = 1$
Thus,
${f_1} = \dfrac{3}{4}\dfrac{v}{L}$
Further, we get
${f_1} = \dfrac{3}{4} \times 6000$
Then, we get
${f_1} = 4500Hz$
Again,
For second overtone,
$n = 2$
Thus,
${f_2} = \dfrac{5}{4}\dfrac{v}{L}$
Further, we get
${f_2} = \dfrac{5}{4} \times 6000$
Then, we get
${f_2} = 7500Hz$
But,
Finding through this process is not a very efficient process.
Thus,
We will apply a generic process.
Now,
${f_n} \leqslant 20000$
As the high range of the audible frequency is $20000Hz$.
Thus,
$\dfrac{{(2n + 1)}}{4}\dfrac{v}{L} \leqslant 20000$
Further, we get
$\dfrac{{\left( {2n + 1} \right)}}{4} \times 6000 \leqslant 20000$
Then, we get
$\left( {2n + 1} \right)1500 \leqslant 20000$
Again, we get
$\left( {2n + 1} \right) \leqslant 13.3$
Then, we get
$2n \leqslant 12.3$
Thus, we get
$n \leqslant 6.15$
Hence,
The maximum number of audible overtones is $6$.

Hence, the correct option is (C).

Note The first harmonic frequency is the fundamental frequency of the wave. Then, from the second harmonics, the frequencies are called the overtone frequencies. Also, all the frequencies including the fundamental and the overtone frequencies are called natural frequencies.