TS EAMCET · Physics · Waves and Sound
A closed pipe is suddenly opened and changed to an open pipe of same length. The fundamental frequency of the resulting open pipe is less than that of 3rd harmonic of the earlier closed pipe by \(55 \mathrm{~Hz}\). Then, the value of fundamental frequency of the closed pipe is
- A \(165 \mathrm{~Hz}\)
- B \(110 \mathrm{~Hz}\)
- C \(55 \mathrm{~Hz}\)
- D \(220 \mathrm{~Hz}\)
Answer & Solution
Correct Answer
(C) \(55 \mathrm{~Hz}\)
Step-by-step Solution
Detailed explanation
Frequency of first mode of vibration \[ n_1=\frac{3 v}{4 /} \] (closed pipe) and frequency of first mode of vibration \[ n_2=\frac{v}{2 l} \] (open pipe) According to the question, \[ n_1-n_2=\frac{3 v}{4 l}-\frac{v}{2 l}=55 \] So \(\quad \frac{3 v-2 v}{4 !}=55\)…
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