a)
[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)
![F=-ky](https://tex.z-dn.net/?f=F%3D-ky)
where
k = 9 is the spring constant
y is the displacement
Also, from Newton's second law of motion, we know that
![F=my''](https://tex.z-dn.net/?f=F%3Dmy%27%27)
where
m = 1 g = 0.001 kg is the mass
y'' is the acceleration
Combining the two equations,
![my''=-ky](https://tex.z-dn.net/?f=my%27%27%3D-ky)
This is a second order differential equation; the solution for y(t) is
![y(t)=A sin(\omega t-\phi)](https://tex.z-dn.net/?f=y%28t%29%3DA%20sin%28%5Comega%20t-%5Cphi%29)
where
A is the amplitude of motion
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
![\phi=0](https://tex.z-dn.net/?f=%5Cphi%3D0)
So the displacement is
![y(t)=A sin(\omega t)](https://tex.z-dn.net/?f=y%28t%29%3DA%20sin%28%5Comega%20t%29)
The velocity of the spring is equal to the derivative of the displacement:
![v(t)=y'(t)=\omega A cos(\omega t)](https://tex.z-dn.net/?f=v%28t%29%3Dy%27%28t%29%3D%5Comega%20A%20cos%28%5Comega%20t%29)
We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,
![v_0=6(0.0254)=0.152 m/s](https://tex.z-dn.net/?f=v_0%3D6%280.0254%29%3D0.152%20m%2Fs)
And since at t = 0, ![cos(\omega t)=1](https://tex.z-dn.net/?f=cos%28%5Comega%20t%29%3D1)
Then we have:
![v_0=\omega A](https://tex.z-dn.net/?f=v_0%3D%5Comega%20A)
From which we find the amplitude:
![A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m](https://tex.z-dn.net/?f=A%3D%5Cfrac%7Bv_0%7D%7B%5Comega%7D%3D%5Cfrac%7B0.152%7D%7B94.9%7D%3D0.0016%20m)
So the solution for the displacement is
[m]
b)
Here we want to find the time t at which the mass returns to equilibrium, so the time t at which
![y=0](https://tex.z-dn.net/?f=y%3D0)
This means that
![sin(\omega t)=0](https://tex.z-dn.net/?f=sin%28%5Comega%20t%29%3D0)
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
![\omega t = \pi](https://tex.z-dn.net/?f=%5Comega%20t%20%3D%20%5Cpi)
which means:
![t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B%5Cpi%7D%7B%5Comega%7D%3D%5Cfrac%7B%5Cpi%7D%7B94.9%7D%3D0.033%20s)
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
![v(t)=\omega A cos(\omega t)](https://tex.z-dn.net/?f=v%28t%29%3D%5Comega%20A%20cos%28%5Comega%20t%29)
where we have
is the angular frequency
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:
![v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s](https://tex.z-dn.net/?f=v%280.033%29%3D%2894.9%29%280.0016%29cos%2894.9%5Ccdot%200.033%29%3D-0.152%20m%2Fs)