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COLOMB DAMPING OR DRY FRICTION DAMPING.
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In case of coulomb damping, frictional force will oppose the motion of mass and it will be always opposite in the direction of velocity.

Consider spring-mass system as shown in following figure.

enter image description here

Case 1: When X is positive.

enter image description here

m$\ddot{x}$ = - kx - F

i.e. m$\ddot{x}$ + kx + F = 0

$\ddot{x} + \frac{k}{m} x + \frac{F}{m} = 0$

$\ddot{x}+ \frac{k}{m} [ x + \frac{F}{k} ] = 0$

Let $ ( x + \frac{F}{K} ) = y$

then $\ddot{x}$ = $\ddot{y}$

Then, the equation of motion of mass can be written as $\ddot{y} + \frac{k}{m} \ y = 0$

i.e. my + ky = 0

Above equation is SHM about y = 0

then above analysis is valid only for that quarter of cycle when X is positive.

then natural frequency of vibration $w_n = \sqrt{\frac1{k}{m}}$

Case 2 : When X is negative:

m$\ddot{x}$+ kx - F = 0

$\therefore \ddot{x} + \frac{k}{m} x - \frac{F}{m} = 0$

$\therefore \ddot{x} + \frac{k}{m} [ x - \frac{F}{k}] = 0$

Let x - \frac{F}{x} = Y and hence $\ddot{x}$ - $\ddot{y}$

The equation of motion can be written as

$\ddot{y} + \frac{k}{m} y = 0$

$\therefore$ my + ky = 0

Above equation is the equation of SHM where it is harmonic about y = 0

i.e. $x - \frac{F}{k} = 0$

Then natural frequency $w_n = \sqrt{k/m}$

Similar analysis can be done for rest of quarters of cycle where natural frequency will be given by equation $w_n = \sqrt{k/m}$

From above analysis, we can say that after some time, mass will come to rest due to energy loss in friction. then displacement v/s time graph is shown below :

enter image description here

Consider points A and B on displacement curve which will represents two successive amplitudes.

Let $x_c$ be the displacement after second quarter of cycle.

Then, by energy principle, we can write,

$\frac{1}{2} kx^2_A - \frac{1}{2} kx_c^2 = F (X_A + X_C)$

$\therefore$ $\frac{1}{2} (X_A + X_C) (X_A - X_C) = F(X_A + X_C)$

$\therefore$ $ X_A - X_C = \frac{2F}{k}$ - - -(1)

Similarly, it can be proved that

$X_C - X_B = \frac{2F}{k}$ - - - - (2)

Adding (1) and (2), we get,

$X_A - X_B = \frac{4F}{k}$

From above equation, it can be said that amplitude lost per cycle is constant and is equal to $\frac{4F}{k}$

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