Answer
u(t) = 0.008Sin(25t) m
Workings
Conditions given:
Mass (m):
m = 250g
=
= 0.25kg
Spring Extension (L):
L = 1.568 cm
=
= 0.01568 m
Let u(t) be the position of the mass at any time, t seconds.
Then,
u(0) = 0 m, the mass is initially in equilibrium.
Velocity = u`(t)
u`(0) = 20cms-1
=
= 0.20ms-1
Finding the spring constant (k)
, where m is the mass in kg, g is the acceleration due to gravity in ms-2 , k is the spring constant in Nm-1 and L is the spring extension in m.
=>
k =
=
=
= 156.25 Nm-1
Differential Equation and solution
When there is no damping and forcing, the equation is given by:
mu`` + ku = 0
The solution to the equation is given by:
u(t) = ASin(w0t) + BCos(w0t), where w0 is the natural frequency of the system and
w02 =
A and B are determined by initial conditions and u(0) = B, u`(0) = Aw0
Since, w02 =
=> w0 =
=
= 25
Therefore,
u(t) = ASin(25t) + BCos(25t)
From initial conditions:
u(0) = ASin(25 × 0) + BCos(25 × 0)
0 = ASin(0) + BCos(0)
B = 0
Thus,
u(t) = ASin(25t) + [0 × Cos(25t)]
u(t) = ASin(25t)
=> u`(t) = 25ACos(25t)
u`(0) = 25ACos(25 ×0)
0.20 = 25ACos(0)
A =
= 0.008
So, u(t) = 0.008Sin(25t) m
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