Mechanics help,find the particular solution and amplitude

[smithy100]

a block P of mass m lies on a smooth plane AB that is inclined at angle (alpha) to the horizontal.The block is attatched to the bottom of the plane,A, by a spring of stiffness 2k and natural length L0.The block is initially released from rest from the equilibrium position.the equilibrium position of the block ,measured from A along the slope is L0-(mg/2k)sin(alpha)
a)find the particular solution of the differential equation that satisfies these initial conditions
b)write down the period and amplitude of the subsequent motion

can some one show me how this is done thanks
how do i find the amplitude?

 

[Nate810]

a) The particular solution of the differential equation that satisfies these initial conditions is given by:y(t) = L0 - (mg/2k)sin(α) + Acos(ωt)where A is the amplitude, and ω is the angular frequency of the motion. The angular frequency is given by:ω = √(2k/m)b) The period of the motion is T = 2π/ω, and the amplitude is given by A.

 

[Moetasim]

a) To find the particular solution, we need to first set up the differential equation that describes the motion of the block. Since the block is attached to a spring, we can use Hooke's law to describe the force acting on the block: F = -kx, where x is the displacement of the block from its equilibrium position.

Next, we can use Newton's second law to relate the force to the acceleration of the block: F = ma. In this case, the acceleration is in the direction of the slope, so we can write the equation as: -kx = mgsin(alpha).

We can rearrange this equation to get the second order differential equation: mx'' + kx = mgsin(alpha), where x'' is the second derivative of x with respect to time.

To find the particular solution, we need to solve this differential equation with the given initial conditions. Since the block is released from rest, we can set x(0) = 0 and x'(0) = 0, where x'(0) is the first derivative of x with respect to time. This gives us the following initial value problem:

mx'' + kx = mgsin(alpha), x(0) = 0, x'(0) = 0

Solving this differential equation, we get the particular solution: x(t) = (mg/k)(sin(alpha)t - (1/2)sin(alpha)^2t^2). This describes the displacement of the block from its equilibrium position as a function of time.

b) To find the period and amplitude of the subsequent motion, we need to use the solution we found in part a). The period of motion is the time it takes for the block to complete one full cycle of oscillation. In this case, since the block is moving along a slope, the period is given by: T = 2pi/sqrt(gsin(alpha)/L0), where L0 is the natural length of the spring.

The amplitude of the motion is the maximum displacement of the block from its equilibrium position. To find this, we can plug in t = T/4 (since the block starts from its equilibrium position and reaches its maximum displacement after one quarter of a cycle) into the particular solution we found in part a). This gives us the amplitude: A = mg/k(sin(alpha)T/4 - (1/2)sin(alpha)^2(T/4)^2).

In