Atube, open at only one end, is cut into two shorter (non-equal) lengths. the piece open at both ends has a fundamental frequency of 425 hz, while the piece open only at one end has a fundamental frequency of 675 hz, both in room temperature air. what is the fundamental frequency of the original tube?

f= 158.6Hz

Explanation

Given Data

f₀=425 HzOpen at both ends has a fundamental frequency

f₁=675 Hzopen only at one end has a fundamental frequency

To find

Fundamental frequency of the original tube

Solution

Fundamental frequency with open ends given as

f = v / (2·L)

where v speed of sound,

and  L is length

And

Fundamental frequency of a tube closed at one end is given as

f = v / (4·L)

For the piece with two open ends:

f₀ = v / (2·L₀)

L₀ = v / (2·f₀)

For the other piece

f₁ = v / (4·L₁)

L₁ = v / (4·f₁)

The fundamental frequency of the original tube is that of a closed end tube of length L₀+ L₁.

f = v/ ( 4·(L₀+ L₁)

f= v/ (4·(v / (2·f₀)+ v / (4·f₁) )

f= 1/ ( 2/f₀+ 1/f₁)  

f= 1/ ( 2 / 425Hz + 1 / 625Hz)

f= 158.6Hz

not moving. pos accel,

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down