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

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.