Question 49: Waves

Question

For a certain organ pipe, the first three resonance frequencies are in the ratio of $$1:3:5$$ respectively. If the frequency of fifth harmonic is $$405 \mathrm{~Hz}$$ and the speed of sound in air is $$324 \mathrm{~ms}^{-1}$$ the length of the organ pipe is _________ $$\mathrm{m}$$.

Accepted Answer

Given that the first three resonance frequencies are in the ratio of $$1:3:5$$, we can express them as follows:
$$f_1 = kf$$
$$f_3 = 3kf$$
$$f_5 = 5kf$$
Where $$k$$ is a constant and $$f_1, f_3$$, and $$f_5$$ are the first, third, and fifth resonance frequencies, respectively. We are given that the frequency of the fifth harmonic is $$405 \mathrm{~Hz}$$, so we can write:
$$f_5 = 5kf = 405 \mathrm{~Hz}$$
Now we can solve for the constant $$k$$:
$$k = \frac{405}{5} = 81 \mathrm{~Hz}$$
We also know that the speed of sound in air is $$v = 324 \mathrm{~ms}^{-1}$$. The relationship between the speed of sound, the frequency, and the wavelength of a standing wave in a closed pipe can be expressed as follows:
$$v = f\lambda$$
Where $$\lambda$$ is the wavelength of the wave. For the first harmonic in a closed pipe, the length of the pipe is equal to one-fourth of the wavelength:
$$L = \frac{1}{4}\lambda$$
We can now substitute the expression for the wavelength in terms of the length into the equation for the speed of sound:
$$v = f_1 \cdot 4L$$
Now, we can substitute the value of $$f_1 = kf = 81 \mathrm{~Hz}$$ and the speed of sound $$v = 324 \mathrm{~ms}^{-1}$$ into the equation:
$$324 = 81 	imes 4L$$
Now we can solve for the length of the organ pipe $$L$$:
$$L = \frac{324}{81 	imes 4} = \frac{324}{324} = 1 \mathrm{~m}$$
The length of the organ pipe is $$1 \mathrm{~m}$$.

Waves

For a certain organ pipe, the first three resonance frequencies are in the ratio of $1:3:5 respectively. If the frequency of fifth harmonic is 405 \mathrm{~Hz} and the speed of sound in air is 324 \mathrm{~ms}^{-1} the length of the organ pipe is _________ \mathrm{m}$.

Integer / Numeric Answer Type