Question

In a closed organ pipe of length \[105\text{ cm}\], standing waves are set up corresponding to the third overtone.
What distance from the closed end, a pressure node is formed?
A.\[5\text{ cm}\]
B.\[15\text{ cm}\]
C.\[25\text{ cm}\]
D.\[30\text{ cm}\]

Answer 1

Hint: In a closed organ pipe only odd harmonics are present, that is, odd multiples of the fundamental frequency.

Formula used: The wavelength \[\lambda \]corresponding nth node is given by
\[\lambda =\dfrac{4L}{(2n-1)}\] where \[L\] is the length of the closed organ pipe.
The position of the nth node is given by
\[x=\dfrac{n\lambda }{2}\]


Complete step by step solution:
The length of the closed organ pipe, \[L=105\text{ cm}\]
The third overtone corresponds to \[n=4\]
So, the wavelength corresponding to the \[n=4\]for an organ pipe of length \[L=105\text{ cm}\] is
\[{{\lambda }_{4}}=\dfrac{4(105\text{ cm)}}{(2(4)-1)}=60\text{ cm}\]
Now, using the value of \[{{\lambda }_{4}}=60\text{ cm}\], the position of the node \[n=4\]is given by
\[{{x}_{4}}=\dfrac{4(60\text{ cm)}}{2}=120\text{ cm}\]
Now, the distance of the pressure node from the closed end is
\[d=120\text{ cm}-105\text{ cm}=15\text{ cm}\]
Therefore the pressure node is formed at a distance \[15\text{ cm}\] from the closed end.
Hence, option B is the correct answer.

Additional information:
 When two waves of same frequency and amplitude moving along the same path but in opposite directions superimpose, the resultant wave is called stationary or standing waves.
Nodes are the points along a standing wave where the amplitude of oscillation of the constituent is zero. In a closed organ pipe one end is open and the other is closed. Nodes are formed at the closed end.
Antinodes are the points along a standing wave where the amplitude of oscillation of the constituents is zero. Antinodes are formed at the open end of a closed organ pipe.


Note: The third overtone corresponds to the seventh harmonic, and is given by
\[{{v}_{3}}=\dfrac{7v}{4L}\], where v denotes the fundamental frequency.