Question

Two open organ pipes 40 cm and 42 cm long found
to give 10 beats in 7 seconds when sounded together,
when each is sounding its fundamental note. Find the
velocity of sound in air.​

Answer 1

Answer:

The required velocity of sound in air is 24 m/s    

Explanation:

Given :

Two open organ pipes 40 cm and 42 cm long found  to give 10 beats in 7 seconds when sounded together,  when each is sounding its fundamental note.

To find :

the velocity of sound in air

Solution :

The fundamental frequency of an open organ pipe is given by,

 ⇒ n = v/2l

where

v denotes the velocity of sound

l denotes the length of the pipe

The two open organ pipes produced 10 beats in 7 seconds. In one second, number of beats produced, Δn = 10/7

 Each pipe is sounding in its fundamental note.

   $\implies \sf \dfrac{v}{2l_1}-\dfrac{v}{2l_2}=\Delta n$

l₁ and l₂ are the lengths of the two open organ pipes.

• l₁ = 40 cm

• l₂ = 42 cm

Substituting the values,

 $\sf \dfrac{v}{2(40)}-\dfrac{v}{2(42)}=\dfrac{10}{7} \\\\ \sf v\bigg( \dfrac{1}{80}-\dfrac{1}{84} \bigg)=\dfrac{10}{7} \\\\ \sf v\bigg( \dfrac{84-80}{80 \times 84}\bigg)=\dfrac{10}{7} \\\\ \sf v\bigg( \dfrac{4}{6720} \bigg)=\dfrac{10}{7} \\\\ \sf v\bigg( \dfrac{1}{1680} \bigg)=\dfrac{10}{7} \\\\ v=\dfrac{10}{7} \times 1680 \\\\ \sf v=10 \times 240 \\\\ \longrightarrow v=2400 \ cm/s \\\\ \implies v=24 \ m/s$

Therefore, the required velocity of sound in air is 24 m/s