Question

The length of an open pipe is half of the length of another closed pipe. When the two pipes are vibrated, four nodes are formed in both the cases. The ratio of the frequencies of the open and the closed pipes is:

• A. $1:1$
• B. $16:7$
• C. $16:3$
• D. $7:3$

Hint:

For an open organ pipe having $N$ nodes, \[ L=N\left(\frac{\lambda}{2}\right). \] For a closed organ pipe having $N$ nodes, \[ L=(2N-1)\left(\frac{\lambda}{4}\right). \] After finding the wavelengths, use \[ f=\frac{v}{\lambda} \] to obtain the frequency ratio quickly without calculating the actual frequencies.

Answer 1

The Correct Option is B

Concept: The formation of stationary waves in organ pipes depends upon the boundary conditions at their ends. In an open organ pipe, both ends are open and therefore antinodes are formed at both ends. In a closed organ pipe, one end is closed and the other end is open; hence a node is formed at the closed end while an antinode is formed at the open end. For an open organ pipe, if $N$ nodes are formed inside the pipe, then the length of the pipe is related to the wavelength by \[ L_o=N\left(\frac{\lambda_o}{2}\right). \] For a closed organ pipe, if $N$ nodes are formed, then the corresponding relation becomes \[ L_c=(2N-1)\left(\frac{\lambda_c}{4}\right). \] The frequency of the sound produced by an air column is related to its wavelength through the fundamental wave equation \[ f=\frac{v}{\lambda}, \] where $v$ is the speed of sound in air. The given problem involves comparing the frequencies of an open and a closed pipe when the number of nodes formed in each case is the same. Therefore, we first determine the wavelengths corresponding to the standing wave patterns and then use the frequency relation to obtain the required ratio.

Step 1: Writing the relation between the lengths of the two pipes. Let the length of the open pipe be $L_o$ and the length of the closed pipe be $L_c$. According to the question, the length of the open pipe is half the length of the closed pipe. Therefore, \[ L_o=\frac{L_c}{2}. \] Rearranging, \[ L_c=2L_o. \] This relation will be used later while comparing the wavelengths of the two pipes.

Step 2: Determining the wavelength corresponding to the open organ pipe. It is given that four nodes are formed in the open organ pipe. Hence, \[ N=4. \] Using the relation for an open organ pipe, \[ L_o=N\left(\frac{\lambda_o}{2}\right), \] we obtain \[ L_o=4\left(\frac{\lambda_o}{2}\right). \] Simplifying, \[ L_o=2\lambda_o. \] Therefore, \[ \lambda_o=\frac{L_o}{2}. \] Now using \[ f_o=\frac{v}{\lambda_o}, \] we get \[ f_o=\frac{v}{L_o/2} =\frac{2v}{L_o}. \] Thus, the frequency of the open organ pipe is \[ f_o=\frac{2v}{L_o}. \]

Step 3: Determining the wavelength corresponding to the closed organ pipe. For the closed organ pipe also, four nodes are formed. Hence, \[ N=4. \] Using the relation \[ L_c=(2N-1)\left(\frac{\lambda_c}{4}\right), \] we obtain \[ L_c=(2\times4-1)\left(\frac{\lambda_c}{4}\right). \] Therefore, \[ L_c=\frac{7\lambda_c}{4}. \] Using the previously obtained relation \[ L_c=2L_o, \] we get \[ 2L_o=\frac{7\lambda_c}{4}. \] Multiplying both sides by $4$, \[ 8L_o=7\lambda_c. \] Hence, \[ \lambda_c=\frac{8L_o}{7}. \] Now using the wave equation, \[ f_c=\frac{v}{\lambda_c}, \] we obtain \[ f_c=\frac{v}{8L_o/7}. \] Therefore, \[ f_c=\frac{7v}{8L_o}. \] Thus, the frequency of the closed organ pipe is \[ f_c=\frac{7v}{8L_o}. \]

Step 4: Calculating the ratio of frequencies of the open and closed pipes. The required ratio is \[ \frac{f_o}{f_c} = \frac{\dfrac{2v}{L_o}} {\dfrac{7v}{8L_o}}. \] Dividing by a fraction is equivalent to multiplying by its reciprocal: \[ \frac{f_o}{f_c} = \frac{2v}{L_o} \times \frac{8L_o}{7v}. \] Cancelling the common factors $v$ and $L_o$, \[ \frac{f_o}{f_c} = \frac{16}{7}. \] Hence, \[ f_o:f_c=16:7. \]

Final Conclusion: Using the standing-wave relations for open and closed organ pipes and the given condition that four nodes are formed in each case, the wavelength of the open pipe is found to be smaller than that of the closed pipe. Since frequency is inversely proportional to wavelength, the frequency of the open pipe is greater. Therefore, the required ratio of frequencies is \[ \boxed{16:7}. \]