91Ó°ÊÓ

1. The ratio of a harmonic frequency to next lower harmonic frequency in pipe A,$\frac{\mathrm{f}}{\mathrm{f}−1}=1.2$
2. The ratio of a harmonic frequency to next lower harmonic frequency in pipe B,$\frac{\mathrm{f}}{\mathrm{f}−1}=1.4$

The number of waves passing a fixed location in a unit of time is referred to as frequency in physics.

We know the resonating frequencies for a pipe with one end and both ends open. From this, we can find the ratios of harmonic frequency to its lower harmonic frequency for both types of pipes. Comparing them with the given ratios we can find the number of open ends in pipes A and B.

A pipe open at both ends resonates at frequencies

$\mathrm{f}={\mathrm{nf}}_0$

Where, $\mathrm{n}=1,2,3,…\mathrm{..}$and ${\mathrm{f}}_0$is fundamental frequency.

Therefore, the ratio of harmonic frequency to its lower harmonic frequency is given as:

$\frac{\mathrm{f}}{\mathrm{f}−1}=\frac{2}{1},\frac{3}{2},….$

It implies that the ratio contains both even and odd numbers.

In the given problem,

$\frac{\mathrm{f}}{\mathrm{f}−1}=1.2=\frac{6}{5}$

The ratio contains both odd and even numbers.

Therefore, number of open ends in pipe A are 2.

A pipe open at one end resonates at frequencies

$\mathrm{f}={\mathrm{nf}}_0$

Where,$\mathrm{n}=1,3,5,…\mathrm{..}$and${\mathrm{f}}_0$ is fundamental frequency.

Therefore, the ratio of harmonic frequency to its lower harmonic frequency is

$\frac{\mathrm{f}}{\mathrm{f}−1}=\frac{3}{1},\frac{5}{3},….$

It implies that the ratio contains only odd numbers in the numerator and denominator.

In the given problem,

$\frac{\mathrm{f}}{\mathrm{f}−1}=1.4=\frac{7}{5}$

The ratio contains only odd numbers.

Therefore, number of open ends in pipe B is 1.