The first harmonic of a closed pipe of length L is 300 Hz. To play the same note on an open pipe the length of the pipe could be

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BlackTom AI · Accepted Answer

The problem links the fundamental frequencies of a closed pipe and an open pipe of related lengths.
First, recall the relationships: for a closed pipe of length L, the fundamental (first harmonic) frequency is f_closed = v/(4L). For an open pipe of length L_open, the fundamental is f_open = v/(2L_open).
From the given, f_closed = 300 Hz, so v = 4 f_closed L = 4 × 300 × L = 1200 L (with v understood as the speed of sound in air).
We want the same note, so set f_open = 300 Hz. Then 300 = v/(2 L_open). Substitute v: 300 = (1200 L)/(2 L_open) → 300 = 600 L / L_open → L_open = 600 L / 300 = 2L.
Now evaluate the options:
- Option "2L": This matches the calculation, so it is the correct choice. The open pipe must be twice as long as the closed pipe to produce the same fundamental frequency.
- Option "25 L": If L_open = 25L, then f_open = v/(2 × 25L) = (1200 L)/(50 L) = 24 Hz, which is far from 300 Hz; this is incorrect.
- Option "L": If L_open = L, then f_open = v/(2L) = (1200 L)/(2L) = 600 Hz; this does not equal 300 Hz, so it’s incorrect.
- Option "5 L": If L_open = 5L, then f_open = v/(2 × 5L) = (1200 L)/(10 L) = 120 Hz; again not 300 Hz, so it’s incorrect.
In summary, only 2L yields the same 300 Hz note on an open pipe.

SPHY001 Practice Quiz - Unit 18 | LA058

The first harmonic of a closed pipe of length L is 300 Hz. To play the same note on an open pipe the length of the pipe could be

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