How does wavelength change depending on whether the pipe is open or closed?

* in a tube with both ends open

        λ = 2L / n                    n = 1, 2, 3,…

* tube with one end open and the other closed.  

         λ = 4L / n                     n = (2n ’+ 1 )

Explanation:

When a sound wave enters a pipe, a resonance process occurs, whereby only some wavelengths can occur.

* in a tube with both ends open

in this case there are maximums at each end, so if the length of the had is L

              λ = 2L 1st harmonic

              λ = 2 L / 2 2nd harmonic

              λ = 2L / 3

              λ = 2L / n                    n = 1, 2, 3,…

* In the case of a tube with one end open and the other closed.

At the open end there is a belly and at the closed end a node

              λ = 4L 1st harmonic

              λ = 4L / 3 3rd harmonic

              λ = 4L / 5

              λ = 4L / n                  n = 1, 3, 5,… odd

                                       n = (2n ’+ 1 )      where n’ are all integers

good looking response

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j