Q121 / 160 · TS EAMCET 2014 · Physics

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
    $110 \mathrm{~Hz}$
  • Correct answer
    C
    $55 \mathrm{~Hz}$
  • D
    $220 \mathrm{~Hz}$

Solution

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 $\quad \frac{3 v-2 v}{4 !}=55$ $$ \Rightarrow \quad \frac{v}{4 l}=55 \mathrm{~Hz} $$

Hence, the correct answer is option: 'C'.

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