User:Virusys/Physics

1 a)

Assuming a dynamic range of 100 dB (the deafening noise of the ancient bell) to 20 dB (general ambient noise floor), we can infer that after 1 minute when the sound dies, the energy of the bell will drop by 80 dB. Therefore, due to the logarithmic nature of decibels, the physicist decay time can easily be calculated by cross multiplying:


 * $$ \frac{80dB_{drop}}{60s} * \frac{2.1dB_{drop}}{\tau} $$

therefore,
 * $$ (80dB_{drop})\tau = (2.1dB_{drop})60s $$
 * $$ \tau = \frac{(2.1dB_{drop})(60s)}{80dB_{drop}} $$
 * $$ \tau = 1.575s $$

Again, this result for the physicist's decay time is assuming that the bell's sound decays exponentially like the simple harmonic oscillator model, and the vibration pattern does not evolve over time (straying from the model).

b)

Given the intrepid physic student's estimation of the bell's power to be 25 Watts, we can assume the initial vibrational energy of the bell using the following equation:


 * $$ P = \frac{E}{\tau} $$

Plugging in the power and our previously derived value for :$$ \tau $$, we get


 * $$ 25W = \frac{E}{1.575s} $$
 * $$ E = (25W)(1.575s) $$
 * $$ E = 39.375 (W)(s) $$

c)


 * $$ E = \frac{1}{2}(2\pi)^2 Mf_0^2A^2 $$
 * $$ \frac{E}{\frac{1}{2}(2\pi)^2 Mf_0^2} = A^2 $$
 * $$ \sqrt{\frac{E}{\frac{1}{2}(2\pi)^2 Mf_0^2}} = A $$

Plugging in our known (estimated) values for E, M, and f:


 * $$ \sqrt{\frac{25W}{\frac{1}{2}(2\pi)^2 (500kg)(100Hz)^2}} = A $$
 * $$ \sqrt{\frac{25W}{98696044.01}} = A $$
 * $$ 0.000503292 = A $$

d) Drawing!

e) Considering that the SHO model used as the basis for these calculations is extremely simplified for the general case, there are numerous factors (perhaps too many to even consider in the real-world) that are unaccounted for. Even though

2 a)

Using a medium-sized cooking pot of approximately 2.5 kg, an estimate of the pitch by striking the underside is 700Hz, which died away after around 0.2 seconds. Using these approximations, we can calculate the Q using the SHO model accounting for exponential decay:


 * $$ Q = 2\pi f_0\tau $$

However, the physicist decay time should first be considered using the technique presented in question 1 a).


 * $$ \frac{40dB}{0.2s} = \frac{2.1dB}{\tau} $$
 * $$ \tau = \frac{(2dB)(0.2s)}{40dB} $$
 * $$ \tau = 0.05s $$

Then using this value to find the Q:


 * $$ Q = 2\pi(700Hz)(0.05s) $$
 * $$ Q = 219.911 $$

2 b)

To relate the mechanical impedance of the cooking pot to its Q and the effective spring constant, the following equation can be derived using the SHO model:
 * $$ Z = \frac{K}{2\pi f_0Q}$$

We get this from:
 * $$ Z = \frac{F}{U} $$


 * $$ = \frac{F}{2\pi fA}$$


 * $$ = \frac{F}{2\pi f} * \frac{K}{QF} $$ because :$$ A = \frac{QF}{K}$$

Therefore,
 * $$ Z = \frac{K}{2\pi f_0Q}$$

The only remaining unknown is K, which can be solved using yet another relation given by the SHO model:


 * $$ f_0 = \frac{1}{2\pi}\sqrt{\frac{K}{M}} $$

Isolating K, we get:


 * $$ K = 4\pi ^2 Mf_0^2 $$
 * $$ K = 4\pi ^2 (2.5kg)(700Hz)^2 $$
 * $$ K = 193444246.261 $$

Plugging in all known values to approximate Z:


 * $$ Z = \frac{193444246.261}{2\pi (700Hz)(219.911)} $$
 * $$ Z \thickapprox 200 $$

c) Going back to the relation
 * $$ Z = \frac{F}{U} $$

We can isolate and estimate the value of U, given that a light tap from a spoon can correspond to a force of about 20 newtons.


 * $$ 200 = \frac{20}{U} $$
 * $$ U = 0.1 m/s $$

Obviously this model does not account for the shape of the pot, and how the force applied to one side of the pot transforms throughout the entire object. The SHO only gives a ballpark estimate for a vibrating surface excited by a force, which is "good enough" in this context.

d) The velocity amplitude is directly proportional to the amplitude A of the SHO model. The reason for this is that a vibrating object with a large amplitude will have to vibrate "faster" to complete each cycle than an object vibrating at the same frequency with a smaller amplitude. Since there is more distance to cover for each cycle of vibration, the velocity amplitude will be higher.

Thus,


 * $$ U = 2\pi f_0 A $$
 * $$ A = \frac{U}{2\pi f_0} $$
 * $$ A = \frac{0.1m/s}{2\pi (700Hz)} $$
 * $$ A = 2.273^-5 m $$

This corresponds to a value of about 0.23 mm. Even though throughout this entire exercise there were many generalizations and simplifications due to using the SHO model, this answer is reasonable. A small amplitude of displacement for striking a cooking pot lightly would intuitively cause it to vibrate a small amount, perhaps too small to notice. A value like 0.23 mm for this displacement would be much more reasonable than anything higher or lower.