If the speed of sound is 340 m/s, the shortest closed pipe which resonates at 218 Hz is:
A) 23 cm
B) 17 cm
C) 39 cm
D) 78 cm
E) 1.56 cm

Question

Quizplus · Accepted Answer

The wavelength of a sound wave can be found using the formula:

wavelength = speed of sound / frequency

In this case, the wavelength would be:

wavelength = 340 m/s / 218 Hz = 1.56 meters

Since the pipe is closed, the wavelength must be equal to four times the length of the pipe divided by three. Therefore:

wavelength = 4L/3

1.56 meters = 4L/3

L = 1.17 meters = 117 cm

However, this is the length of the pipe that would resonate at the fundamental frequency (the first harmonic). The problem asks for the shortest closed pipe that resonates at 218 Hz, which would be the second harmonic. The second harmonic has a wavelength of twice the fundamental frequency, so:

wavelength = 2 * 1.56 meters = 3.12 meters

Now we can solve for the length of the pipe:

3.12 meters = 4L/3

L = 2.34 meters = 234 cm

But again, this is the length for the fundamental frequency. We want to find the shortest pipe that resonates at the second harmonic frequency, so we need to divide this length by two:

234 cm / 2 = 117 cm

So the answer is C, 39 cm.

If the Speed of Sound Is 340 M/s, the Shortest