Question 41: Waves

Question

Two open organ pipes of lengths $$60 \mathrm{~cm}$$ and $$90 \mathrm{~cm}$$ resonate at $$6^{	ext {th }}$$ and $$5^{	ext {th }}$$ harmonics respectively. The difference of frequencies for the given modes is _________ $$\mathrm{Hz}$$. (Velocity of sound in air $$=333 \mathrm{~m} / \mathrm{s}$$)

Accepted Answer

To solve this problem, let's first understand how the harmonics of open organ pipes work. For an open organ pipe, the harmonics are given by the formula:
$$f_n = n \frac{v}{2L}$$
where:
$$f_n$$ is the frequency of the nth harmonic,
$$n$$ is the harmonic number (an integer),
$$v$$ is the speed of sound in air,
$$L$$ is the length of the organ pipe.
Given that the speeds of sound in air $$v = 333 \, 	ext{m/s}$$, and the lengths of the two open organ pipes are $$60 \, 	ext{cm} = 0.60 \, 	ext{m}$$ and $$90 \, 	ext{cm} = 0.90 \, 	ext{m}$$, we can calculate the frequencies of the 6th harmonic for the 60 cm pipe and the 5th harmonic for the 90 cm pipe.
For the 60 cm pipe at the 6th harmonic ($$n = 6$$):
$$f_{6,60} = 6 \frac{333}{2 	imes 0.60} = 6 	imes \frac{333}{1.2} = 6 	imes 277.5 = 1665 \, 	ext{Hz}$$
For the 90 cm pipe at the 5th harmonic ($$n = 5$$):
$$f_{5,90} = 5 \frac{333}{2 	imes 0.90} = 5 	imes \frac{333}{1.8} = 5 	imes 185 = 925 \, 	ext{Hz}$$
The difference in frequencies between these two modes is:
$$\Delta f = f_{6,60} - f_{5,90} = 1665 \, 	ext{Hz} - 925 \, 	ext{Hz} = 740 \, 	ext{Hz}$$
Therefore, the difference of frequencies for the given modes is $$740 \, 	ext{Hz}$$.

Waves

Two open organ pipes of lengths $60 \mathrm{~cm} and 90 \mathrm{~cm} resonate at 6^{\text {th }} and 5^{\text {th }} harmonics respectively. The difference of frequencies for the given modes is _________ \mathrm{Hz}. (Velocity of sound in air =333 \mathrm{~m} / \mathrm{s}$)

Integer / Numeric Answer Type