Free Vibration of a Mass Spring System Without Damping

Fig.1 gives the different stages of a freely vibrating system without damping. In Fig.1(c) the displacement ‘z’ from the position of static equilibrium at a certain time ‘t’ is shown. The static deflection is given by,

$k=\dfrac{W}{z_s}$ ———-(1)

Fig.1 (f) shows the free body diagram of the system. Applying Newton’s second law we get,

$W-(W + k\times z) = m \times \dfrac {d^2z}{dt^2}$ ———- (2)

Or, $-k\times z = m\times \dfrac {d^2z}{dt^2}$ ———- (3)

Or, $m \times \dfrac {d^2z}{dt^2}+k\times z = 0$ ———- (4)

Or, $\dfrac{d^2z}{dt^2}+\dfrac {k}{m}\times z = 0$ ———- (5)

Put $z = e^{\lambda t}$ ———- (6)

Then $\dfrac{d^2z}{dt^2} = \lambda ^2\times$ $e^{\lambda t}$ ———- (7)

By substation the value from equation (6) and (7) in equation (5),

$\lambda^2 \times e^{\lambda t} + \dfrac{k}{m}\times e^{\lambda t}$ = 0 ———- (8)

Or, $\lambda ^2 + \dfrac{k}{m}$ = 0 ———- (9)

Or, $\lambda ^2 = -\dfrac{k}{m}$ ———- (10)

Or, $\lambda = \pm\sqrt{-\dfrac{k}{m}} = \pm i\sqrt{\dfrac{k}{m}}$ ———- (11)

By substitution $\lambda$ from equation (11) into equation (6) we have,

$z = e^{\pm I \sqrt{\dfrac{k}{m}\times t}}$ ———- (12)

$z = A\times e^{\pm I \sqrt{\dfrac{k}{m}\times t}}$ + B\times $z = e^{-\pm I \sqrt{\dfrac{k}{m}\times t}}$ ———- (13)

Substituting the value of e in equation (13) we get,

$z = A Cos\left(\sqrt{\dfrac{k}{m}}\times t\right) + Sin\left(\sqrt{\dfrac{k}{m}}\times t\right) + B Cos\left(\sqrt{\dfrac{k}{m}}\times t\right) - Sin\left(\sqrt{\dfrac{k}{m}}\times t\right)$

$= (A + B)Cos\left(\sqrt{\dfrac{k}{m}}\times t\right) + (A - B)Sin\left(\sqrt{\dfrac{k}{m}}\times t\right)$

$= C_1 Cos\left(\sqrt{\dfrac{k}{m}}\times t\right) + C_2 Sin\left(\sqrt{\dfrac{k}{m}}\times t\right)$

$z = C_1{Cos(\omega_n\times t)} + C_2{Sin(\omega_n \times t)}$ ———- (14)

Where, $\omega_n = \sqrt{\dfrac{k}{m}} = 2\times\pi\times f_n$ ———- (15)

and $\omega_n$ is called the undamped natural angular velocity of the vibrating system. f_n is the undamped natural frequency. Hence, we can write the equation for natural frequency as,

$f_n = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$ Cycles/sec ———- (16)

And the time period is given by

$T = 2\pi\sqrt{\dfrac{m}{k}}$ ———- (17)

By substituting, $m = \dfrac{w}{g}$ and $z_s = \dfrac{w}{k} in equations (16) and (17)</p> <p>[latex]f_n = \dfrac{1}{2\pi}\sqrt{\dfrac{g}{z_s}}$ ---------- (18)

And $T = 2\pi\sqrt{\dfrac{z_s}{g}}$ ---------- (19)

The above equations clearly shows that the un-damped single degree of freedom vibrating system is harmonic at a natural frequency $f_n$ . The amplitude of vibration can be obtained by putting boundary conditions. At t = 0, z = z₀ and $\dfrac{dz}{dt} = v_0$ , hence by substituting this condition in equation (14) we get,