Posted by shasha on Thursday, March 10, 2011 at 10:03pm.
Question 1:
Experimentally it is found that a 6 kg weight stretches a certain spring 6 cm. If the weight is pulled 4 cm below the equilibrium position and released:
a. Set up the differential equation and associated conditions describing the motion.
b. Find the position of the weight as a function of time.
c. Find the amplitude, period, and frequency of motion
d. Determine the position, velocity and acceleration of the weight 0.5s after it has been released.
Question 2:
A 0.25 kg mass is horizontally attached to a spring with a stiffness 4 N/m. The damping constant b for the system is 1 Nsec/m. If the mass is displaced 0.5 m to the left and given an initial velocity of 1 m/sec to the left, find the equation of motion. What is the maximum displacement that the mass will attain?

calculus  MathMate, Friday, March 11, 2011 at 3:50pm
Question 1:
The standard differential equation of motion is:
mx"+Bx'+kx=f(t) ....(1)
where
x" is the second derivative with respect to time of displacement, x
x' is the first derivative
m=mass
B=damping
k=spring constant
f(t)=applied external force
For the given case,
m=6 kg
B=0
k=mg/h=6*9.81/0.06=981 N/m
f(t)=0
Initial conditions:
x(0)=0.04 (note: we use the convention positive downwards)
x'(0)=0
So substitute the quantities to equation 1:
6x"+981x=0 .... (1a)
Auxiliary equation:
6m²+981=0
m=±12.79i => α=a=0, β=b=12.79
So solution to equation (1a)
x=C1 cos(bt)+C2 sin(bt)
Apply initial conditions:
x(0)=0.04 = C1(1)+C2(0) => C1=0.04
x'(t)=C1*b*sin(bt)+C2*b*cos(bt)
x'(0)=0 = C1(0)+C2*b(1) => C2=0/b=0
Therefore C1=0.04, C2=0, or
x(t)=0.04cos(12.79t)
x'(t) = d(x(t))/dt = 12.79*0.04sin(12.79t)=0.511sin(12.79t)
I will leave it to you to answer parts (c) and (d).
Q2 has been answered previously, see:
http://www.jiskha.com/display.cgi?id=1299811520

calculus  shasha, Saturday, March 12, 2011 at 6:46am
thanxxxxx

calculus  shasha, Saturday, March 12, 2011 at 7:12am
thanxxxxx...what is * stand for??

calculus  MathMate, Saturday, March 12, 2011 at 7:17am
If f(x) is a function, then f'(x)=dy/dx.
If the independent variable is known or understood (often either x or t), then
y'=dy/dx or dy/dt, depending on the context.
In this case, the independent variable understood is t, so x(t) is the function, x' stands for x'(t)=dx/dt, and x" stands for x"(t)=d²x/dt².

calculus  farah, Monday, May 20, 2013 at 9:02am
i still don't get it the answer for question 1, c and d..can u give me the answer please..
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