Physics Question please HELP

Question

An organ pipe open at one end and closed at the other is 30 cm long. If the temperature in the church is 30 °C, what is the fundamental frequency and the first overtone of the pipe?

Answer 1

Pipes or tubes that are closed at one end and open at the other end will produce harmonics or resonant sounds when the pipe has the following pattern of lengths: L =  (1/4)λ, or (3/4)λ, or (5/4)λ, or (7/4)λ, or (odd#/4)λ...

The fundamental occurs at shortest length, L = (1/4)λ. The first overtone occurs at the next harmonic where L = (3/4)λ.

To find the frequency we use the equation: v = f • λ thus f = (v/λ)

GIVEN:

L = 30 cm = 0.30 m

speed of sound = 343.4 at 20°C and changes by 0.6 m/s for every degree above or below 20. So at 30°C, a difference of 10°C, the speed would be faster by (0.6 x 10) = 6 m/s faster.

v_s= 349.4 m/s

λ_fundamental = (4 x L) = 4 x 0.3 m = 1.2 m

λ_{first overtone} = (4/3)(L) = (4/3)(0.3m) = 0.4 m

UNKNOWN:

f_fundamental = (v_s / λ_fundamental) = (349.4 m/s / 1.2 m) = 291.2 Hz

f_{first overtone} = (v_s / λ_{first overtone}) = (349.4 m/s / 0.4 m) = 873.5 Hz

 

Another pattern that emerges is that each successive overtone frequency is an odd multiple of the fundamental frequency. So the f of the first overtone is 3x the frequency of the fundamental. 

( 873.5/291.2) = 3. The second overtone is 5x the fundamental frequency, the next is 7x, then 9x....