Question

A mass of 1 g is set in motion from its equilibrium position with an initial velocity of 6in/sec, with no damping and a spring constant of k = 9. Find (a) the solution y for the displacement of the spring, (b) the time when the mass first returns to equilibrium (i.e. displacement y = 0), (c) it's velocity at the time you found in (b). slader

Answer 1

a) $y(t)=0.0016 sin(94.9t)$ [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)

$F=-ky$

where

k = 9 is the spring constant

y is the displacement

Also, from Newton's second law of motion, we know that

$F=my''$

where

m = 1 g = 0.001 kg is the mass

y'' is the acceleration

Combining the two equations,

$my''=-ky$

This is a second order differential equation; the solution for y(t) is

$y(t)=A sin(\omega t-\phi)$

where

A is the amplitude of motion

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{9}{0.001}}=94.9 rad/s$ 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

$\phi=0$

So the displacement is

$y(t)=A sin(\omega t)$

The velocity of the spring is equal to the derivative of the displacement:

$v(t)=y'(t)=\omega A cos(\omega t)$

We know that at t = 0, the initial velocity is 6 in/s; since 1 in = 2.54 cm = 0.0254 m,

$v_0=6(0.0254)=0.152 m/s$

And since at t = 0, $cos(\omega t)=1$

Then we have:

$v_0=\omega A$

From which we find the amplitude:

$A=\frac{v_0}{\omega}=\frac{0.152}{94.9}=0.0016 m$

So the solution for the displacement is

$y(t)=0.0016 sin(94.9t)$ [m]

b)

Here we want to find the time t at which the mass returns to equilibrium, so the time t at which

$y=0$

This means that

$sin(\omega t)=0$

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

$\omega t = \pi$

which means:

$t=\frac{\pi}{\omega}=\frac{\pi}{94.9}=0.033 s$

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

$v(t)=\omega A cos(\omega t)$

where we have

$\omega=94.9 rad/s$ is the angular frequency

$A=0.0016 m$ 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:

$v(0.033)=(94.9)(0.0016)cos(94.9\cdot 0.033)=-0.152 m/s$