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Two springs of negligible mass have the same natural lengt (50cm) and the same spring...

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kibauu | (Level 2) eNoter

Posted June 11, 2013 at 6:44 AM via web

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Two springs of negligible mass have the same natural lengt (50cm) and the same spring constant. In figure 1, the two springs are joined together and then are suspendly vertically with lower end also fixed in place. In figure 2, a small object A (mass 500 g)is attached at the point where the springs are joined, object A descends 20 cm and come to rest.

A is pulled downward and released gently. How much time elapses from release until A reaches its maximum height?

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valentin68 | College Teacher | (Level 3) Associate Educator

Posted September 12, 2013 at 1:09 PM (Answer #1)

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See below for the diagram for the forces acting on mass m.

The elastic constant is

`2*k*x =m*g`

`k=(m*g)/(2*x) = (0.5*9.81)/(2*0.2)=12.2625 N/m`

Total force is

`F = m*g -kx -kx = mg -2k*x`

equation of motion for mass m is

`m*a = F`  or equivalent `m*(d^2x)/(dt^2) +2k*x =m*g`

rewritten the equation becomes

`(d^2x)/(dt^2) +((2*k)/m)*x =g`

If we define `omega = sqrt((2k)/m)`

the solutions of the above second order differential equation are

`x(t) = X_0*cos(omega*t) + "constant"`

where `X_0` is the elongation at t=0 and the value of the constant is determined by entering the solution x(t) into the equation.

Now we can determine the frequency of the oscillations

`F = omega/(2*pi) = 1/(2*pi) *sqrt((2*k)/m)`

and the period

`T = 1/F = 2*pi*sqrt(m/(2*k))`

The time for A to reach the maximum height is just

`t = T/2 = pi*sqrt(m/(2*k)) =pi*sqrt(0.5/(2*12.2625)) =0.4486 seconds`

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