A 250-g object hangs from a spring and oscillates with an amplitude of
5.42 cm. If the spring constant is 48.0 N/m, determine the acceleration of the object when the displacement is 4.27 cm [down].

Question

A 250-g object hangs from a spring and oscillates with an amplitude of 
5.42 cm. If the spring constant is 48.0 N/m, determine the acceleration of the object when the displacement is 4.27 cm [down].

homeworksucks · Answer

I did that and got an acceleration of ~8,19 m/s^2

homeworksucks · Answer

*8.19

anonymous · Answer

$$F=ma ; F=kx$$
Set those equal to each other to find the acceleration

homeworksucks · Answer

My only problem is how to find the velocity, I got a max velocity of 0.75 m/s which sounds weird given the acceleration

homeworksucks · Answer

I used the formula $$v=A \sqrt{\frac{k}{m}}$$

loser66 · Answer

I want to confirm something. Is it ODE for oscillation harmonic function  or physics?

homeworksucks · Answer

Physics, I posted it to Physics, but didn't have any luck

loser66 · Answer

ok, @ybarrap  It's physics, friend, help him please

loser66 · Answer

we have $\omega^2 = \dfrac {k}{m}= \dfrac{48.0}{0.25}= 192$
and acceleration $\dfrac{d^2x}{dt^2}= -\omega^2 A cos (\omega t+ \phi)$

loser66 · Answer

@ybarrap  is here, wait for him, hehehe. you will have a neat explanation.

ybarrap · Answer

|dw:1386534100699:dw|
$$
x(t)=A\cos(\omega t + \phi)\
A=.0542\text{ m}\
\text{Acceleration = }\cfrac{d^2x(t)}{dt^2}=-A\omega^2\cos(\omega t + \phi)\qquad(\star)
$$
$x(0)=0$, so $\phi=\cfrac{\pi}{2}$
We need to know when $x(t)=.0427m$.

So, 

$$
x(t)=.0427=-.0542\times\cfrac{48}{.250}\cos(\sqrt{\cfrac{48}{.250}} t + \cfrac{\pi}{2})
$$
After you solve for $t$, insert back into $\star$ above and you are done.

Make sense?