Question

The velocity of sound in air at room temperature is \[350\,{\text{m/s}}\]. An air column \[35\,{\text{cm}}\] is in length. Find the frequency of the third overtone in a pipe, when it is (a) closed at one end (b) open at both ends.

Answer 1

Hint: Use the formula for velocity of a wave in terms of frequency of the wave. Recall the values of wavelengths of the waves for the third overtone when an organ pipe is closed at one end and open at both the ends. Substitute these values in the formula and determine the required values of frequencies.

Formula used:
The velocity \[v\] of a wave is given by
\[v = n\lambda \] …… (1)
Here, \[n\] is frequency of the wave and \[\lambda \] is wavelength of the wave.

Complete step by step answer:
(a) We have given that the velocity of the sound wave in the organ pipe is \[350\,{\text{m/s}}\].
\[v = 350\,{\text{m/s}}\]
The length of the organ pipe is \[35\,{\text{cm}}\].
\[L = 35\,{\text{cm}}\]
The wavelength \[{\lambda _3}\] of the wave in an organ pipe closed at one end for third overtone is
\[{\lambda _3} = \dfrac{{4L}}{3}\]
Rewrite equation (1) for the velocity of the sound wave in the organ pipe closed at one end for third overtone.
\[v = {n_3}{\lambda _3}\]
Here, \[{n_3}\] is the frequency for the third overtone in an organ pipe closed at one end.
Rearrange the above equation for \[{n_3}\].
\[{n_3} = \dfrac{v}{{{\lambda _3}}}\]
Substitute \[\dfrac{{4L}}{3}\] for \[{\lambda _3}\] in the above equation.
\[{n_3} = \dfrac{v}{{\dfrac{{4L}}{3}}}\]
\[ \Rightarrow {n_3} = \dfrac{{3v}}{{4L}}\]

Substitute \[350\,{\text{m/s}}\] for \[v\] and \[35\,{\text{cm}}\] for \[L\] in the above equation.
\[ \Rightarrow {n_3} = \dfrac{{3\left( {350\,{\text{m/s}}} \right)}}{{4\left( {35\,{\text{cm}}} \right)}}\]
\[ \Rightarrow {n_3} = \dfrac{{3\left( {350\,{\text{m/s}}} \right)}}{{4\left( {35\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)}}\]
\[ \Rightarrow {n_3} = \dfrac{{3\left( {350\,{\text{m/s}}} \right)}}{{4\left( {0.35\,{\text{m}}} \right)}}\]
\[ \Rightarrow {n_3} = 750\,{\text{Hz}}\]

Hence, the frequency of the third overtone in an organ pipe closed at one end is \[750\,{\text{Hz}}\].

(b) The wavelength \[{\lambda _3}\] of the wave in an organ pipe open at both ends for third overtone is \[{\lambda _3} = \dfrac{{2L}}{3}\]
Rewrite equation (1) for the velocity of the sound wave in the organ pipe open at both ends end for third overtone.
\[v = {n_3}{\lambda _3}\]
Here, \[{n_3}\] is the frequency for the third overtone in the organ pipe open at both ends.

Rearrange the above equation for \[{n_3}\].
\[{n_3} = \dfrac{v}{{{\lambda _3}}}\]
Substitute \[\dfrac{{2L}}{3}\] for \[{\lambda _3}\] in the above equation.
\[{n_3} = \dfrac{v}{{\dfrac{{2L}}{3}}}\]
\[ \Rightarrow {n_3} = \dfrac{{3v}}{{2L}}\]
Substitute \[350\,{\text{m/s}}\] for \[v\] and \[35\,{\text{cm}}\] for \[L\] in the above equation.
\[ \Rightarrow {n_3} = \dfrac{{3\left( {350\,{\text{m/s}}} \right)}}{{2\left( {35\,{\text{cm}}} \right)}}\]
\[ \Rightarrow {n_3} = \dfrac{{3\left( {350\,{\text{m/s}}} \right)}}{{2\left( {35\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)}}\]
\[ \Rightarrow {n_3} = \dfrac{{3\left( {350\,{\text{m/s}}} \right)}}{{2\left( {0.35\,{\text{m}}} \right)}}\]
\[ \therefore {n_3} = 1500\,{\text{Hz}}\]

Hence, the frequency of the third overtone in an organ pipe open at both ends is \[1500\,{\text{Hz}}\].

Note: The students should be careful while using the values of the wavelengths of the waves for the third overtone in the organ pipe closed at one end and open at both ends. If these values are taken incorrect then the final value of the frequency will also be incorrect.