Question

The wavelength (a) of second harmonic in a closed pipe and the length of
pipe (l) are related as of
(A) lamda= 4l
(B) lamda= 4l/3
(C)lamda = 2l
(D) lamda= 2l/3​

Answer 1

Answer:

4l/3

Explanation:

length l=λ/2 +λ/4 =3λ/4

For 3rd harmonic there will be a node and an antinode.

distance between 2 nodes are λ/2 and dist. b/w node and an antinode is λ/4.

therefore length of the pipe isgiven above.