A tuba creates a 4th harmonic of
1 answer:
Answer:
92.33Hz
Explanation:
A tuba is considered as having one open end and the other end is closed. Then:
f_n = n×v / 4L
so here
Where n = 4
f_4 = 116.5 Hz
116.5 Hz = nv/4L
116.5 Hz = 4v/4L
4×v / 4L = v / L
116.5 Hz = v/L............1
Let's assume the speed of sound is 343 m/s. Then substituting 343 m/s for v in equation 1
116.5 Hz = v/L
116.5 Hz = 343 m/s/L
Making L the subject
116.5 Hz × L = 343 m/s
L = 343m/s / 116.5Hz
= 2.944 m
Add 0.721 m to the length, and
f_4 = 343m/s / (2.944+0.721)m
f_4 = 343m/s ÷ 3.715
= 92.328 Hz
Approximately = 92.33Hz
Hence the new frequency for the 4th harmonic is 92.33Hz
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