A pendulum located where $g = 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }$ has a 0.500 kg bob and a 4.00 Hz oscillation frequency.If it is released from a point 0.150 m above the lowest point it reaches,the number of quanta in this oscillation is A)2.77 * 10³². B)1.11 * 10³³. C)4.87 *10 ^- ³⁴. D)0.184. E)4.00.

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Quizplus · Accepted Answer

The energy of the pendulum is:$$E = \frac{1}{2}mv^2 + mgh = \frac{1}{2}m\omega^2A^2 + mgA(1-\cos\theta)$$where $m$ is the mass of the bob, $v$ is the velocity of the bob, $\omega$ is the angular frequency, $A$ is the amplitude, $g$ is the acceleration due to gravity, $h$ is the height of the bob above the lowest point, and $\theta$ is the angle of the pendulum from the vertical.At the lowest point, $\theta = 0$ and $v = \omega A$, so the energy is:$$E = \frac{1}{2}m\omega^2A^2 + mgA$$At the highest point, $\theta = \pi$ and $v = 0$, so the energy is:$$E = mgA(1-\cos\pi) = 2mgA$$The number of quanta in the oscillation is:$$n = \frac{E}{hf}$$where $h$ is the Planck constant and $f$ is the frequency.Substituting the given values, we get:$$n = \frac{\frac{1}{2}m\omega^2A^2 + mgA}{hf} = \frac{\frac{1}{2}(0.500 \text{ kg})(2\pi \times 4.00 \text{ Hz})^2(0.150 \text{ m})^2 + (0.500 \text{ kg})(9.80 \text{ m/s}^2)(0.150 \text{ m})}{(6.626 \times 10^{-34} \text{ J s})(4.00 \text{ Hz})} = 1.11 \times 10^{33}$$Therefore, the correct answer is B.

A Pendulum Located Where g=9.80ms2g = 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }g=9.80s2m​

A Pendulum Located Where $g = 9.80 \frac { \mathrm { m } } { \mathrm { s } ^ { 2 } }$