show that for a clósed organ pipe only odd harmonics are present wheras for an open organ pipe all the harmonics are present​

For a closed organ pipe, the point at the closed end is always a node, 'N' while the open end is always an antinode, 'A'

Therefore, for a closed organ pipe, we only have;

NA = N↔A = First harmonics = f₁

NANA = NA↔AN↔ NA = Third harmonics, = f₃ = 3·f₁

NANANA = NA↔AN ↔ NA ↔ AN ↔ NA = f₅ = 5·f₁

It can also be shown that for a closed organ pipe, only f₇, f₉, f₁₁, ... are present

Therefore, a closed organ pipe has only odd harmonics present

For an open organ pipe, we have;

ANA = First harmonics = f₁

ANANA = ANA ↔ ANA = Second harmonics = f₂ = 2·f₁

ANANANA = ANA ↔ ANA ↔ ANA = Third harmonics = f₃ = 3·f₁

Similar relationship exist for f₄, f₅, f₆...

Therefore, for an open organ pipe, all harmonics, (multiples of the fundamental, f₁) are present

Explanation:

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.

none, its a machine