A crystal oscillator used for measuring the thickness of evaporated metal films

Question

A crystal oscillator used for measuring the thickness of evaporated metal films is located in a vacuum chamber and has an oscillation frequency of 5 kHz. (FYl 1 kHz = 1000 Hz). When air is introduced into the vacuum chamber, the oscillation amplitude decreases to 50% of its initial amplitude in 10 seconds. Calculate the corresponding quality factor of the oscillator (after air has been admitted into the chamber). Also calculate the total number of cycles of oscillation that have occurred when the oscillator reaches 10% of its initial amplitude. - Quality factor of a lightly damped SHO.

Explanation / Answer

Quality factor, also called Q factor is defined as
Q = w * Maximum Energy / power loss

so for the given system f = 5*10^3 Hz
so, w = 2*pi*f = 31.41*10^3 rad/s

Maximum energy stored in the osscilator = 0.5kA^2 [ where k is spring constant, A is its amplitude]
energy loss = 0.5k(A/2)^2 = 0.5kA^2/4
power loss = energy loss / 10 seconds

so Q = 31.41*10^3 *0.5kA^2 *4*10/0.5*kA^2 = 31.41*10^3 *4*10 = 1.2566*10^6

now, as Q factor is very high, f remains unchanged
so time period of one osscilation, T = 1/f
T = 1/5*10^3 = 0.0002 s
so in 10 sec, number of time periods = 10/0.0002 = 50,000 osscilations

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