Question

Two open organ pipes A and B of same length 0.5 m are blown with air at different temperatures so as to produce 6 beats per second. A is blown with air at 27 C and B is blown with air at higher temperature t C. the value of t is?(assume the fundamental mode for both)

Answer 1

$\displaystyle\Large\boxed {\sf {\quad t=37.7\quad}}$

Solution:-

The fundamental frequency of an open pipe of length L and velocity of sound through it, v, is given by,

$\displaystyle\longrightarrow\sf{\nu=\dfrac {v}{2L}}$

Given two open pipes A and B having same length 0.5 m.

Let,

• $\displaystyle\sf {\nu_{A}=}$ fundamental frequency of the pipe A.

• $\displaystyle\sf {\nu_{B}=}$ fundamental frequency of the pipe B.

• $\displaystyle\sf {v_A=}$ velocity of sound in pipe A.

• $\displaystyle\sf {v_B=}$ velocity of sound in pipe B.

Assume that $\displaystyle\sf {\nu_B\ \textgreater\ \nu_A.}$ Because given that the temperature at pipe B (t °C) is higher than that at pipe A.

Given that the no. of beats produced is 6. So,

$\displaystyle\longrightarrow\sf{\nu_B-\nu_A=6}$

$\displaystyle\longrightarrow\sf{\dfrac {v_B}{2\times0.5}-\dfrac {v_A}{2\times 0.5}=6}$

$\displaystyle\longrightarrow\sf{v_B-v_A=6}$

$\displaystyle\longrightarrow\sf{v_B=v_A+6\quad\quad\dots (1)}$

Well, the relation between the velocity of sound v and temperature T is as follows.

$\displaystyle\longrightarrow\sf{v=\sqrt{\dfrac {\gamma RT}{M}}\quad\quad\dots (2)}$

where,

• $\displaystyle\sf {\gamma=\dfrac {C_p}{C_v},}$ known as adiabatic constant.

• R = universal gas constant

• M = molecular mass

From this we get the following proportionality,

$\displaystyle\longrightarrow\sf{v\propto\sqrt{T}}$

And therefore,

$\displaystyle\longrightarrow\sf{\dfrac {v_B}{v_A}=\sqrt{\dfrac {T_B}{T_A}}\quad\quad\dots (3)}$

Here,

• $\displaystyle\sf {T_A=27^oC=300\ K}$

• $\displaystyle\sf {T_B=t^oC=(t+273)\ K}$

Then from (3),

$\displaystyle\longrightarrow\sf{\dfrac {v_B}{v_A}=\sqrt{\dfrac {t+273}{300}}}$

$\displaystyle\longrightarrow\sf{\dfrac {\left (v_B\right)^2}{\left (v_A\right)^2}=\dfrac {t+273}{300}}$

From (1),

$\displaystyle\longrightarrow\sf{\dfrac {\left (v_A+6\right)^2}{\left (v_A\right)^2}=\dfrac {t+273}{300}}$

$\displaystyle\longrightarrow\sf{t=300\cdot\dfrac {\left (v_A+6\right)^2}{\left (v_A\right)^2}-273\quad\quad\dots (4)}$

But we have to find out the value of $\displaystyle\sf {v_A.}$

In the case of air,

• $\displaystyle\sf {\gamma=1.4}$

• $\displaystyle\sf {M=0.03\ kg\ mol^{-1}}$

And we know that,

• $\displaystyle\sf {R=8.314\ J\ K^{-1}\ mol^{-1}}$

Then from (2),

$\displaystyle\longrightarrow\sf{v_A=\sqrt{\dfrac {1.4\times8.314\times300}{0.03}}}$

$\displaystyle\longrightarrow\sf{v_A=341\ m\ s^{-1}}$

This is the normal speed of sound too. $\displaystyle\sf {(T_A=300\ K}$ is actually the normal atmospheric temperature.)

So (4) becomes,

$\displaystyle\longrightarrow\sf{t=300\cdot\dfrac {\left (341+6\right)^2}{\left (341\right)^2}-273}$

$\displaystyle\longrightarrow\sf{t=300\cdot\dfrac {\left (347\right)^2}{\left (341\right)^2}-273}$

$\displaystyle\longrightarrow\sf{t=300\cdot\dfrac {120409}{116281}-273}$

$\displaystyle\longrightarrow\sf {\underline {\underline {t=37.7}}}$