Originally Posted by **galactus** Hey all
My first query on the forum. I am trying to relearn a little physics. Here is a problem I am unsure of. **Here we have a mass, M1, attached to a spring (pardon my sloppy 'Paint' **
spring) of unstretched length d and pulled by a constant force of
$\displaystyle F_{2}=m_{2}g$.
Suppose the system is in equilibrium when x=L. Is L>d or L<d?. If L and d are known, what is the spring constant k?.
If the system is at rest in the position x=L and the mass M2 is suddenly removed. Say we cut the wire or string holding them. Then what is the period and amplitude of oscillation of M1?.
I would think that L>d.
Equilibrium is $\displaystyle y=\frac{-m}{k}g$ due to M2. Therefore, x=L=y.
That is, the mass M2 moves the mass M1 horizontally the same amount as it displaces when it is hanging down.
Since at equilibrium, the forces balance and we have $\displaystyle \frac{d^{2}y}{dt^{2}}=0$, then $\displaystyle y=\frac{-m}{k}g$
Could we then solve for $\displaystyle k=\frac{g}{y}m$?.
Period of oscillation is $\displaystyle T=2{\pi}\sqrt{\frac{m}{k}}$
and amplitude A in $\displaystyle x=Asin({\omega}t+{\phi})$
How can I relate these and solve the problem.
I have many thoughts but am unsure about them. |

You've got most of it except that you are looking for x(t), not y(t) (especially after the string has been cut!) You have the equilibrium position correct as

$\displaystyle x = \frac{mg}{k} + d$

(In order to write this in terms of x you need to include the unstretched length d to this.)

You have the period of the motion correct.

For the amplitude, think of this in terms of an energy argument. Can the (ideal) spring go any further beyond d than mg/k? Then that must be your amplitude.

-Dan