The standard differential equation of motion is:
x" is the second derivative with respect to time of displacement, x
x' is the first derivative
f(t)=applied external force
For the given case,
x(0)=0.04 (note: we use the convention positive downwards)
So substitute the quantities to equation 1:
6x"+981x=0 .... (1a)
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'(0)=0 = C1(0)+C2*b(1) => C2=0/b=0
Therefore C1=0.04, C2=0, or
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:
thanxxxxx...what is * stand for??
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.
i still dont get it the answer for question 1, c and d..can u give me the answer please..
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