Physics question? On Resonance

[drake]

Sorry if i posted in the wrong area, but I'm new here and I couldn't find a suitable area to post this:

The third resonance length of a closed air column is resonating to a tuning fork at 95cm. Determine the first and second resonance lengths.

 

[PlanetMaster]

Sorry about late reply!

Here's your answer:
1st resonance:
λ/4 = 76/4 = 19cm

2nd resonance:
3λ/4 = (76x3)/4 = 57cm

3rd resonance:
5λ/4 = (76x5)/4 = 95cm

 

[anzaar]

Your question requires some more information to get the answer. I explain the topic here .
For the first resonance length the formula would be
c+l1 =λ/4 ; where c is end correction.
For 2nd resonance length
c+l2 =3λ/4
and so on hence for nth resonance length the formula would be
c+ln =(2n-1)λ/4 ; n=1,2,3…………..
of
c+ln =(2n-1)v/4f where v=f λ
In the form of frequency when length is fixed
The fundamental note f0 , will be l= λ0 /4
Implies l= v/4f0
Implies f0= v/4l
For harder blow
l=3/4 λ1 ; v=f1 λ1
f1 =(3/4)v/l
Implies that f1 =3f0
Similarly, f2 =5f0
f0, 3f0,5f0..............

Odd harmonics
You need either to find end correction etc.