Answer:
2.94 cm
20 cm
Explanation:
Solution:-
- We are to consider a mass-spring system that initially undergoes free vibrations with a time period ( T ) given to be 1.25s.
- The natural frequency (w_n) of the free vibrations undergone by the system can be determined as:
[tex]w_n = \frac{2\pi }{T} \\\\w_n = \frac{2\pi }{1.25}\\\\w_n = 5.02654\frac{rad}{s}[/tex]
- The system is later excited with a base displacement of the form:
y ( t ) = Y*sin ( w*t )
- The upper end of the spring is made to move with the following displacement function:
y ( t ) = 5*sin (2π*t)
Where,
Y : The amplitude of excitation = 5
w : The excited frequency = 2*π
- The amount of height raised of the center of mass in the first 0.4 seconds can be determined from the excitation displacement ( y ( t ) ).
- The simple plugging of the t = 0.4 s in the displacement of function " y ( t ) ", hence:
[tex]y ( 0.4 ) = 5*sin (2\pi *0.4)\\\\y ( 0.4 ) = 5*sin (2.51327)\\\\y ( 0.4 ) = 2.94 cm[/tex]
Answer: There will be a rise of 2.94 cm of the mass in the first 0.4 seconds.
- The equation of motion for the base excitation of mass-spring-damper system is given as follows:
[tex]m*\frac{d^2x}{dt^2} + c*\frac{dx}{dt} + kx = k*y(t) + c*\frac{dy}{dt}[/tex]
Where,
m: The mass of the object ( point mass )
c : The viscous damping coefficient
k: The spring stiffness constant
x : The absolute motion of mass ( free vibration + excitation )
- For undamped system ( ζ = 0 ) i.e the damping coefficient ( c ) is zero. The complete solution of the mass-spring system is given in the form:
[tex]m*\frac{d^2x}{dt^2} + kx = k*y(t) \\\\\frac{d^2x}{dt^2} + w_n^2*x = w_n^2*y(t)[/tex]
Where,
[tex]w_n^2 = \frac{k}{m}[/tex]
- The steady solution of an undamped mass-spring system is given in the form:
[tex]x_s_s ( t ) = X_o sin ( w*t )[/tex]
Where,
X_o : The amplitude of the mass for steady-state vibration.
- The general amplitude ( X_o ) for a damped system is given by the relation:
[tex]X_o = Y*\sqrt{\frac{1 + ( 2*p*r )^2}{( 1 - r ) ^2 + ( 2*p*r )^2} }[/tex]
Where,
Y: The amplitude of exciting displacement ( y ( t ) )
p = ζ ( Damping ratio constant )
[tex]r = \frac{w}{w_n} = \frac{2*\pi }{\frac{2\pi }{T} } = T= 1.25[/tex]
- The undamped system ( ζ = 0 ) i.e the damping coefficient ( c ) is zero. The amplitude ( X_o ) of a steady state response is given as:
[tex]X_o = Y*\sqrt{\frac{1 + ( 2*0*r )^2}{( 1 - r ) ^2 + ( 2*0*r )^2} }\\\\X_o = Y*\sqrt{\frac{1 }{( 1 - r ) ^2 } }\\\\X_o = Y*\frac{1}{1-r}[/tex]
- Plug in the value of ( r ) and evaluate the amplitude of steady state response of the mass-spring vibration.
[tex]X_o = 5*\sqrt{\frac{1}{(1-1.25)^2} } \\\\X_o = 5*\sqrt{\frac{1}{0.0625} }\\\\X_o = 20 cm[/tex]
Answer: The steady state amplitude of the mass vibrations is ( X_o ) = 20 cm
