a) [tex]y(t)=0.0016 sin(94.9t)[/tex] [m]
b) 0.033 s
c) -0.152 m/s
Step-by-step explanation:
a)
The force acting on the mass-spring system is (restoring force)
[tex]F=-ky[/tex]
where
k = 9 is the spring constant
y is the displacement
Also, from Newton's second law of motion, we know that
[tex]F=my''[/tex]
where
m = 1 g = 0.001 kg is the mass
y'' is the acceleration
Combining the two equations,
[tex]my''=-ky[/tex]
This is a second order differential equation; the solution for y(t) is
[tex]y(t)=A sin(\omega t-\phi)[/tex]
where
A is the amplitude of motion
[tex]\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{9}{0.001}}=94.9 rad/s[/tex] is the angular frequency
The spring starts its motion from its equilibrium position, this means that y=0 when t=0; therefore, the phase shift must be
[tex]\phi=0[/tex]
So the displacement is
[tex]y(t)=A sin(\omega t)[/tex]
The velocity of the spring is equal to the derivative of the displacement:
[tex]v(t)=y'(t)=\omega A cos(\omega t)[/tex]
We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,
[tex]v_0=6(0.0254)=0.152 m/s[/tex]
And since at t = 0, [tex]cos(\omega t)=1[/tex]
Then we have:
[tex]v_0=\omega A[/tex]
From which we find the amplitude:
[tex]A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m[/tex]
So the solution for the displacement is
[tex]y(t)=0.0016 sin(94.9t)[/tex] [m]
b)
Here we want to find the time t at which the mass returns to equilibrium, so the time t at which
[tex]y=0[/tex]
This means that
[tex]sin(\omega t)=0[/tex]
We know already that the first time at which this occurs is
t = 0
Which is the beginning of the motion.
The next occurence of y = 0 is instead when
[tex]\omega t = \pi[/tex]
which means:
[tex]t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s[/tex]
c)
As said in part a), the velocity of the mass-spring system at time t is given by the derivative of the displacement, so
[tex]v(t)=\omega A cos(\omega t)[/tex]
where we have
[tex]\omega=94.9 rad/s[/tex] is the angular frequency
[tex]A=0.0016 m[/tex] is the amplitude of motion
t is the time
Here we want to find the velocity of the mass when the time is that calculated in part b):
t = 0.033 s
Substituting into the equation, we find:
[tex]v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s[/tex]
