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What is the fundamental frequency of a 0.672-m-long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic?

a) 254 Hz; (b) 508 Hz
b) 508 Hz; (b) 1016 Hz
c) 254 Hz; (b) 1270 Hz
d) 508 Hz; (b) 762 Hz

User Baadshah
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Final answer:

The fundamental frequency of the tube is 254 Hz and the frequency of its second harmonic is 508 Hz.

Step-by-step explanation:

The fundamental frequency of a tube open at both ends can be calculated using the formula:

f = v / (2L)

where f is the frequency, v is the speed of sound, and L is the length of the tube.

In this case, the length of the tube is 0.672 m and the speed of sound is 344 m/s.

Plugging in these values, we get:

f = 344 / (2 * 0.672) = 254 Hz

The second harmonic is twice the fundamental frequency, so:

second harmonic = 2 * 254 = 508 Hz

User Simon Groenewolt
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