Harmonic motion whose damping is not related to velocity

[fdsadsfa]

Homework Statement

A block of mass m rests on a horizontal table and is attached to a spring of force constant k. The coefficient of friction between the block and the table is mu. For this problem we will assume that the coefficients of kinetic and static friction are equal. Let the equilibrium position of the mass be x = 0. The mass is moved to the position x = +A, stretching the spring, and then released.

1 Apply Newton's 2nd law to the block to obtain an equation for its acceleration for the first half cycle of its motion, i.e. the part of its motion where it moves from x = +A to x < 0 and (momentarily) stops. Show that the resulting equation can be written d^2x'/dt^2 = -omega2 * x', where x' = x - x0 and x0 = mu*m*g/k. Write the expression for position of the block, x(t), for the first half cycle (be sure to express omega, the angular frequency, in terms of the constants given in the problem statement). What is the smallest value of x that the mass reaches at the end of this first half cycle?

2 Repeat the above for the second half cycle, i.e. wherein the block moves from its maximum negative position to its (new) maximum positive position. First show that the differential equation for the block's acceleration can be written d^2x''/dt^2 = -omega2*x'' where this time x'' = x + x0. Next, match the amplitude for the beginning of this half cycle with the amplitude at the end of the last one. Write the expression for the position of the block, x(t), for the second half cycle.

3 Make a graph of the motion of the block for the first 5 half cycles of the motion in the case where A = 10.5*x0. Plot the position of the block normalized to x0 as a function of the fractional period, T = 2*Pi/omega (i.e. plot x(t)/x0 vs t/T).

Homework Equations

The Attempt at a Solution

I got a half way down to the problem 1., i.e., i got an equation for d^x/dt^2
which is eqaul to -k/m(x-(mu*m*g)/k).
But I don't really get how to find a generic equation for position x(t).
If it were a mere simple harmonic motion, it might have been easy; x(t) = Acos(wt).
However, with a daming factor (mu*m*g)/k, I don't see any way to figure out a formula for position x(t)...
I'll be looking forward to your suggestions or ideas on this problem. Thank you!

 

[Redbelly98]

Homework Statement

A block of mass m rests on a horizontal table and is attached to a spring of force constant k. The coefficient of friction between the block and the table is mu. For this problem we will assume that the coefficients of kinetic and static friction are equal. Let the equilibrium position of the mass be x = 0. The mass is moved to the position x = +A, stretching the spring, and then released.

1 Apply Newton's 2nd law to the block to obtain an equation for its acceleration for the first half cycle of its motion, i.e. the part of its motion where it moves from x = +A to x < 0 and (momentarily) stops. Show that the resulting equation can be written d^2x'/dt^2 = -omega2 * x', where x' = x - x0 and x0 = mu*m*g/k. Write the expression for position of the block, x(t), for the first half cycle (be sure to express omega, the angular frequency, in terms of the constants given in the problem statement). What is the smallest value of x that the mass reaches at the end of this first half cycle?

The Attempt at a Solution

I got a half way down to the problem 1., i.e., i got an equation for d^x/dt^2
which is eqaul to -k/m(x-(mu*m*g)/k).
But I don't really get how to find a generic equation for position x(t).
If it were a mere simple harmonic motion, it might have been easy; x(t) = Acos(wt).
However, with a daming factor (mu*m*g)/k, I don't see any way to figure out a formula for position x(t)...
I'll be looking forward to your suggestions or ideas on this problem. Thank you!

How about rewriting your equation for d^x/dt^2 in terms of x', rather than x, as suggested in the problem statement?

 

[thomate1]

The equation for position x(t) can be solved by using the general solution for a damped harmonic oscillator, which is given by:

x(t) = Ae^(-beta*t)cos(omega*t + phi)

where A is the amplitude, beta is the damping coefficient, omega is the angular frequency, and phi is the phase angle. In this case, we can rewrite the equation as:

x(t) = Ae^(-mu*m*g*t/k)cos(omega*t + phi)

To find the values of A and phi, we can use the initial conditions given in the problem. At t = 0, x = A, so we have:

A = Ae^(0)cos(phi)

which gives us A = A. And at t = 0, dx/dt = 0, so we have:

0 = -A*mu*m*g/k*sin(phi)

which gives us phi = 0.

Therefore, the equation for position x(t) becomes:

x(t) = Ae^(-mu*m*g*t/k)cos(omega*t)

To find the smallest value of x, we can set the derivative of x(t) with respect to t equal to 0. This will give us the point where the block momentarily stops before moving in the opposite direction. So we have:

dx/dt = -A*(mu*m*g/k)*e^(-mu*m*g*t/k)*sin(omega*t) = 0

which gives us sin(omega*t) = 0, or omega*t = n*pi, where n is an integer. The smallest value of x will occur at the first time this happens, so we have:

omega*t = pi

and since we know that omega = sqrt(k/m), we can plug this in to get:

sqrt(k/m)*t = pi

and solving for t, we get:

t = pi*sqrt(m/k)

Substituting this value of t back into the equation for x(t), we get:

x(t) = Ae^(-mu*m*g*pi/k)cos(omega*t)

and since e^(-mu*m*g*pi/k) will be a constant, we can rewrite this as:

x(t) = A'cos(omega*t)

where A' is just a different constant. This shows that the smallest value of x will occur when cos(omega*t) is equal to -1, or when omega*t = (2n + 1