A 15-kg mass,attached to a massless spring whose force constant is 2 500 N/m,has an amplitude of 4 cm.Assuming the energy is quantized,find the quantum number of the system,n,if E_n = nhf. A)1.5 * 10³³ B)3.0 *10³³ C)4.5 *10³³ D)5.4 *10³³ E)1.0 *10³³

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The energy of the system is given by:$$E_n = nhf$$where:* $n$ is the quantum number* $h$ is the Planck constant* $f$ is the frequency of the systemThe frequency of the system is given by:$$f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}$$where:* $k$ is the force constant* $m$ is the massSubstituting the given values into the equation for the frequency, we get:$$f = \frac{1}{2\pi}\sqrt{\frac{2500 	ext{ N/m}}{15 	ext{ kg}}} = 2.83 	ext{ Hz}$$Substituting the given values into the equation for the energy, we get:$$E_n = nhf = n(6.626 	imes 10^{-34} 	ext{ J s})(2.83 	ext{ Hz}) = 1.87 	imes 10^{-33} n 	ext{ J}$$The amplitude of the system is given by:$$A = \sqrt{\frac{2E_n}{k}}$$Substituting the given values into the equation for the amplitude, we get:$$A = \sqrt{\frac{2(1.87 	imes 10^{-33} n 	ext{ J})}{2500 	ext{ N/m}}} = 1.13 	imes 10^{-16} \sqrt{n} 	ext{ m}$$The given amplitude is 4 cm, which is equal to $4 	imes 10^{-2} 	ext{ m}$. Setting this equal to the equation for the amplitude, we get:$$4 	imes 10^{-2} 	ext{ m} = 1.13 	imes 10^{-16} \sqrt{n} 	ext{ m}$$Solving for $n$, we get:$$n = \left(\frac{4 	imes 10^{-2} 	ext{ m}}{1.13 	imes 10^{-16} 	ext{ m}}ight)^2 = 1.26 	imes 10^{33}$$Therefore, the quantum number of the system is $n = 1.26 	imes 10^{33}$.

A 15-Kg Mass,attached to a Massless Spring Whose Force Constant