MHT CET · Physics · Waves and Sound
The first overtone of the vibrating air column of a closed pipe has the same frequency as the second overtone of an open pipe \(1.5 \mathrm{~m}\) Jung. The length of the closed pipe is
- A 1.0 m
- B 0.75 m
- C 0.5 m
- D 1.25 m
Answer & Solution
Correct Answer
(B) 0.75 m
Step-by-step Solution
Detailed explanation
Frequency is given by \(\frac{v}{\lambda}\), where \(v\) is the speed of the sound and \(\lambda\) the length of the pipe.
For the second overtone of an open pipe:
\(\begin{aligned}
& \frac{3}{2} \lambda=L_O \\
& \therefore f_O=\frac{3 v}{2 L_O}
\end{aligned}\)
For the first overtone of the closed pipe:
\(\begin{aligned}
& \frac{3}{4} \lambda=L_C \\
& \therefore f_C=\frac{3 v}{4 L_C}
\end{aligned}\)
If they have the same frequency:
\(\frac{3 v}{2 L_O}=\frac{3 v}{4 L_C}\)
\(\Rightarrow L_C=\frac{L_O}{2}=\frac{1.5}{2} \mathrm{~m}\)
For the second overtone of an open pipe:
\(\begin{aligned}
& \frac{3}{2} \lambda=L_O \\
& \therefore f_O=\frac{3 v}{2 L_O}
\end{aligned}\)
For the first overtone of the closed pipe:
\(\begin{aligned}
& \frac{3}{4} \lambda=L_C \\
& \therefore f_C=\frac{3 v}{4 L_C}
\end{aligned}\)
If they have the same frequency:
\(\frac{3 v}{2 L_O}=\frac{3 v}{4 L_C}\)
\(\Rightarrow L_C=\frac{L_O}{2}=\frac{1.5}{2} \mathrm{~m}\)
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