A pipe of length 0.85 m, closed at one…

A pipe of length 0.85 m, closed at one end and open at the other, is tested with tuning forks of various frequencies at an air temperature of 300 K…

One or more options correct

A pipe of length

0.85\ \mathrm{m}

, closed at one end and open at the other, is tested with tuning forks of various frequencies at an air temperature of

300\ \mathrm{K}

, where the speed of sound is

340\ \mathrm{m\,s^{-1}}

. Neglect end correction. Which of the following statements is/are correct?

A. A tuning fork of frequency $700\ \mathrm{Hz}$ sets the air column into resonance.
B. A tuning fork of frequency $600\ \mathrm{Hz}$ sets the air column into resonance.
C. If the closed end is opened (pipe open at both ends), the fundamental frequency becomes $50\ \mathrm{Hz}$ .
D. If the air temperature is raised to $600\ \mathrm{K}$ , the fundamental frequency of the original closed pipe becomes $200\ \mathrm{Hz}$ .

Show answer and solution

Answer: A

For a pipe closed at one end the resonant frequencies are

f_n=(2n+1)\dfrac{v}{4L}

. Here

\dfrac{v}{4L}=\dfrac{340}{3.4}=100\ \mathrm{Hz}

, so the allowed frequencies are

100,300,500,700,\dots\ \mathrm{Hz}

: odd multiples of

100\ \mathrm{Hz}

only. A

700\ \mathrm{Hz}

fork is the seventh harmonic and resonates (A true), but

600\ \mathrm{Hz}

is an even multiple and cannot (B false). If both ends are open the fundamental is

v/2L=200\ \mathrm{Hz}

, i.e. it doubles, not halves to

50\ \mathrm{Hz}

(C false). Since

v\propto\sqrt T

, doubling the absolute temperature multiplies the speed by

\sqrt2

, so the fundamental becomes

100\sqrt2\approx141\ \mathrm{Hz}

rather than

200\ \mathrm{Hz}

(D false).

Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake