Answer:
The description for the given question is described in the explanation section below.
Explanation:
[tex]x = A \ cos(wt + phi)[/tex]...(equation 1)
[tex]dx/dt = v = - Aw \ sin( wt + phi)[/tex]...(equation 2)
at t = 0,
x = 10 &,
v = 0,
From equation 1, we get
⇒ [tex]A = 10 \ cos(phi)[/tex]...(equation 3)
⇒ [tex]0 = - 10 w \ sin(phi)[/tex]
⇒ [tex]phi=0[/tex]
By using this value in equation 3, we get
[tex]A = 10 \ cos(0)[/tex]
[tex]A = 10 \ cm[/tex]
Now from equation 2, we get
[tex]dv/dt = a = -w \ 2A \ cos(wt)[/tex]...(equation 4)
As we know,
[tex]Acceleration = \frac{F}{m}[/tex]
at t = 0,
a = -kA/m
= w2 A
w = root(k/m)
= 15 rad/sec
(b)...
From equation 2, we get
[tex]Vmax = wA[/tex]
[tex]=1.5 \ m/sec[/tex]
(c)...
From equation 4, we get
[tex]a = - 152\times 0.1 cos (15\times 0.15)[/tex]
[tex]=-14.15 \ m/sec2[/tex]
