An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 hz what is the distance from the center of the pipe to the nearest antinode a. 13 cm b. 9.8 cm c. 6.5 cm d. 3.2 cm e. 0 cm

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

So the distance will be 6.5 cm

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

answerdmitry mendeleyev made a workable classification of the elements