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,

principle of free vibration without damping

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\timese^{\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. fn 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,

C_2 = \dfrac{v_0}{\omega_n} and C_1 = z_0

Therefore equation (14) may be expressed as,

z = \dfrac{v_0}{\omega_n}{Sin(\omega_n\times t)} + z_0{Cos(\omega_n\times t)} ---------- (20)

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