Question

The trunk of a elephant may extend up to 2.1 m. It acts much like an organ pipe only open at one end when the elephant blows air through it.Calculate the fundamental frequency of the sound the elephant can create with its trunkIf the elephant blows even harder, the next harmonic may be sounded. What is the frequency of the first overtone?

Answer 1

f1: 40.83Hz

f2: 81.66Hz

Step-by-step explanation:

To find the fundamental frequency you use the formula for semi-closed tubes:

`f_n=(nv_s)/(4L)`

for n=1 you obtain the fundamental:

`f_1=((1)(343m/s))/(4(2.1m))=40.83Hz`

the fundamental is 43.83Hz

the first overtone is obtained for n=2:

`f_2=((2)(343m/s))/(4(2.1m))=81.66Hz`

the first overtone is 81.66Hz

Answer 2

f = 40.5 Hz, f = 121.4 Hz

Step-by-step explanation:

As they indicate that we approach the elephant's trunk as a system with one side closed and the other open. On the closed side there is a node and the open side has a maximum, so we have ¼ wavelength at this distance

L = ¼ λ

For higher order resonance we have

L = 1 (¼ λ ) fundamental

L = 3 (¼ λ ) 1st harmonic

L = 5 (¼ λ ) 3 harmonic

Wave speed is related to wavelength and frequency

v = λ f

f = v /λ

Let's apply the initial equations to this problem

Fundamental frequency

λ = 4L

f = v / 4L

The speed of sound is v = 340 m / s

f = 340 / (4 2.1)

f = 40.5 Hz

The first harmonic

λ = 4/3 L

f= v 3 / 4L

f = 3 340 / (4 2.1)

f = 121.4 Hz