Physics: 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

In 83.33 seconds, the runner should complete 1000 metres...

For small amplitudes of oscillation the motion of

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14.01.2022

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,

$\Delta E = hv$

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

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

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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

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