Question

A tuba creates a 4th harmonic of

Answer 1

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