# Thread: mechanics help , a block is on an incline need to find the period and amplitude of mo

1. ## mechanics help , a block is on an incline need to find the period and amplitude of mo

a block P of mass m lies on a smooth plane AB that is inclined at angle of (alpha) to the horizontal. the block is attatched to the bottom of the plane,A, by a spring of stiffness 2k and natural length Lo . a second model spring of stifness 3k and natural length 2Lo is now attatched to the block, and its other end is fixed at B at a distance 5Lo up the plane from A.
a) the block is initially released from rest from the equilibrium position. i need to find the particular solution of the differential equation that satisfies these initial conditions.
b)i need to write down the period and amplitude of the subsequent motion

if some one could please go through it with me step by step i would be very thankfull .

2. Originally Posted by longy
a block P of mass m lies on a smooth plane AB that is inclined at angle of (alpha) to the horizontal. the block is attatched to the bottom of the plane,A, by a spring of stiffness 2k and natural length Lo . a second model spring of stifness 3k and natural length 2Lo is now attatched to the block, and its other end is fixed at B at a distance 5Lo up the plane from A.
a) the block is initially released from rest from the equilibrium position. i need to find the particular solution of the differential equation that satisfies these initial conditions.
b)i need to write down the period and amplitude of the subsequent motion

if some one could please go through it with me step by step i would be very thankfull .
I don't have much time, but I'll give you an outline.

Let the origin for x be at the $L_0$ position for spring A. Then the "stretch" of spring A is x and the "stretch" of spring B is 2L_0 - x. I am calling down the slope positive.

Set up your Newton's second to find a. Remember we have a weight term to consider here depending on the angle, but the spring forces don't. So we get
$a = \frac{1}{m} \left [ mg~sin(\theta) + 2k(L_0 + x) - 3k(2L_0 - x) \right ]$

So
$\frac{d^2x}{dt^2} - \frac{5k}{m} \cdot x = g~sin(\theta) - \frac{4kL_0}{m}$
(Double check each of these to make sure I'm not making any mistakes. As I said, I'm running through this quickly.)

The right hand side is just a constant, so the particular solution will be in the form of x = C.

To get the period etc. solve the homogeneous equation
$\frac{d^2x}{dt^2} - \frac{5k}{m} \cdot x = 0$

This will be a sine function of the form $A~sin(kx + \phi )$ which you can get your period and amplitude information from.

-Dan