how would i solve these problem?

Question

1) How many overtones are present within the audible range for a 2.33 m long organ pipe at 12°C if it is open?

2) How many are present if it is closed?

Answer 1

Lets look at the equations
For open pipe
fn(open)=n(v/(2L)
fn - frequency at a n-th harmonic
n - harmonic (overtone) fundamental n=1
L- length of the pipe

Similarly for closed pipe
fn(closed)=n(v/(4L)

Air speed as a function of temperature is
v=331.5 + sqrt(1+t/273.15) or approximately

v=331.5 + 0.06t t in degrees Celsius.

V(12 deg C)=331.5 + 0.6 x 12= 338.7 m/s

So let's do the fundamental first
f1=(1)338.7/(2 x 2.33)=72.68Hz
Is audible range 20KHz?
then n=f(max)2L/v=20,000 x 2 x 2.33/ 338.7=275
n=275! wow
if you lose it it will double!
to 550 overtones....?
These are theoretical values so take it with a grain of salt.