Question

For small amplitudes of oscillation the motion of a pendulum is simple harmonic. Consider a pendulum with a period of 0.550 s Find the ground-level energy. Express your result in joules Find the ground-level energy. Express your result in election volts

Answer 1

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

$T = 0.55s$

Therefore the frequency will be the inverse of the period and would be given as

$f= \frac{1}{T}$

$f = \frac{1}{0.55}$

$f = 1.82s^{-1}$

The ground state energy of the pendulum is,

$E = \frac{1}{2} hv$

$E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})$

$E = 6.03*10^{-34}J$

The ground state energy in eV,

$E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})$

$E = 3.8*10^{-15}eV$

The energy difference between adjacent energy levels,

$\Delta E = hv$

$\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)$

$\Delta E = 12.1*10^{-34}J$