To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to
[tex]T = 0.55s[/tex]
Therefore the frequency will be the inverse of the period and would be given as
[tex]f= \frac{1}{T}[/tex]
[tex]f = \frac{1}{0.55}[/tex]
[tex]f = 1.82s^{-1}[/tex]
The ground state energy of the pendulum is,
[tex]E = \frac{1}{2} hv[/tex]
[tex]E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})[/tex]
[tex]E = 6.03*10^{-34}J[/tex]
The ground state energy in eV,
[tex]E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})[/tex]
[tex]E = 3.8*10^{-15}eV[/tex]
The energy difference between adjacent energy levels,
[tex]\Delta E = hv[/tex]
[tex]\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)[/tex]
[tex]\Delta E = 12.1*10^{-34}J[/tex]
