Question

An organ pipe ${{P}_{1}}$ closed at one end and vibrating in its first overtone pipe ${{P}_{2}}$ open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of lengths of ${{P}_{1}}$ to that of ${{P}_{2}}$ is
$A.\,\dfrac{3}{8}$
$B.\,\dfrac{1}{3}$
$C.\,\dfrac{1}{2}$
$D.\,\dfrac{8}{3}$

Answer 1

Firstly we will find the lengths of the pipes using the given condition of the pipes, that is, open or closed and the vibrations produced by them. Then, we will represent the wavelengths in terms of the frequencies using the suitable formula. Finally, we will equate these frequency values to find the ratio of the lengths of the pipes.

Formula used:

& f=\dfrac{v}{\lambda } \\\ \end{aligned}$$ **Complete step by step answer:** From the data, we have the data as follows. An organ pipe $${{P}_{1}}$$ closed at one end is vibrating in its first overtone. As this pipe is closed at one end and is vibrating in its first overtone, the mathematical representation of the same is given as follows. $${{L}_{1}}=\dfrac{3{{\lambda }_{1}}}{4}$$ …… (1) Now, we will make use of the formula that relates the parameters such as the frequency and wavelength to represent the above equation in terms of the frequency. $${{f}_{1}}=\dfrac{v}{{{\lambda }_{1}}}$$ $$\Rightarrow {{\lambda }_{1}}=\dfrac{v}{{{f}_{1}}}$$ …… (2) Equate the equation (2) in (1), so, we get, $$\begin{aligned} & {{L}_{1}}=\dfrac{3}{4}\dfrac{v}{{{f}_{1}}} \\\ & \Rightarrow {{f}_{1}}=\dfrac{3}{4}\dfrac{v}{{{L}_{1}}} \\\ \end{aligned}$$ …… (3) An organ pipe $${{P}_{2}}$$ open at both ends is vibrating in its third overtone. As this pipe is open at both the ends and is vibrating in its third overtone, the mathematical representation of the same is given as follows. $${{L}_{2}}=\dfrac{4{{\lambda }_{2}}}{2}$$ …… (4) Now, we will make use of the formula that relates the parameters such as the frequency and wavelength to represent the above equation in terms of the frequency. $${{f}_{2}}=\dfrac{v}{{{\lambda }_{2}}}$$ $$\Rightarrow {{\lambda }_{2}}=\dfrac{v}{{{f}_{2}}}$$ …… (5) Equate the equation (5) in (4), so, we get, $$\begin{aligned} & {{L}_{2}}=\dfrac{4}{2}\dfrac{v}{{{f}_{2}}} \\\ & \Rightarrow {{f}_{2}}=\dfrac{4}{2}\dfrac{v}{{{L}_{2}}} \\\ \end{aligned}$$ …… (6) From the given, we have the condition that the vibrations produced by both the pipes are in resonance with a given tuning fork. Thus, the mathematical representation of the same is given as follows. $${{f}_{1}}={{f}_{2}}$$ Substitute the equations (3) and (6) in the above equation. $$\dfrac{3}{4}\dfrac{v}{{{L}_{1}}}=\dfrac{4}{2}\dfrac{v}{{{L}_{2}}}$$ Cancel out the common terms and represent the equation in terms of the ratio of the lengths of the pipes. $$\begin{aligned} & \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{3}{4}\times \dfrac{2}{4} \\\ & \Rightarrow \dfrac{{{L}_{1}}}{{{L}_{2}}}=\dfrac{3}{8} \\\ \end{aligned}$$ $$\therefore $$ The value of the ratio of lengths of $${{P}_{1}}$$ to that of $${{P}_{2}}$$ is $$\dfrac{3}{8}$$ . **So, the correct answer is “Option A”.** **Note:** The mathematical representation that relates the parameters such as, first overtone, third overtone and the closed or open end pipes should be known to solve this type of problems. The relation between the frequency and the wavelength should also be known.