Simple Harmonic Motion and friction

[henryc09]

Homework Statement

Mass m moves along a line on a rough table and is attached on either side to a stretched spring (both have equal spring constants k). The coefficients of static and sliding friction between mass and the table are equal with value $$\mu$$

(a) show in the abscence of friciton the particle executes SHM with angular frequency$$\omega$$ $$\sqrt{2k/m}$$

(b) Now include the effect of friction. Suppose particle is released at time t=0 with positive displacement Xo from equilibrium. Describe initial motion when (i)2kXo> $$\mu$$mg and (ii) when 2kXo =< $$\mu$$mg

(c) For case (i) write down the equation of satisfied by the displacement x of the mass as long as it remains moving. Verify it is satisfied by a solution of the form:

x(t)=Acos(wt) + Bsin(wt) + C

and find the values of A, B and C for the data given.

(d) Find the time t1 and position x1 at which the particle next comes to rest.

Homework Equations

The Attempt at a Solution

Been a while since I last did S.H.M, but so far I have without friction total force on object will be -2kx which is equal to -mw²x so i can show the first bit.

For part b I said that when 2kXo > $$\mu$$mg it will oscillate with lower frequency and it's amplitude will gradually decrease, when 2kX0 =< $$\mu$$mg it will remain stationary.

Part C is where I'm stuck. I think -mw²x= - 2kx + $$\mu$$mg but not sure how to show the solution and also how to go about working out the constants. Any point in the right direction would be appreciated.

 

[anathema]

Hi there,

For part C, you are correct that the equation of motion is given by -mw^2x = -2kx + μmg. To solve for the constants A, B, and C, we can use the initial conditions given in part b. For case (i), when 2kXo > μmg, the initial conditions are x(0) = Xo and x'(0) = 0 (since the particle is released from rest). Plugging these into the general solution x(t) = Acos(wt) + Bsin(wt) + C, we get:

Xo = A + C and 0 = -Awsin(wt) + Bwcos(wt)

From the first equation, we can solve for A in terms of Xo and C: A = Xo - C. Plugging this into the second equation and using the trig identity sin^2(wt) + cos^2(wt) = 1, we get:

0 = -w(Xo - C)sin(wt) + Bwcos(wt)

Since this equation must hold for all values of t, we can set the coefficients of sin(wt) and cos(wt) equal to 0:

-w(Xo - C) = 0 and Bw = 0

Solving these equations, we get B = 0 and C = Xo. Therefore, the solution for case (i) is:

x(t) = (Xo - A)cos(wt) + Xo

For case (ii), when 2kXo =< μmg, the initial conditions are x(0) = Xo and x'(0) = 0 (since the particle is released from rest). Plugging these into the general solution x(t) = Acos(wt) + Bsin(wt) + C, we get:

Xo = A + C and 0 = -Awsin(wt) + Bwcos(wt)

Using the same steps as above, we get A = 0 and C = Xo. Therefore, the solution for case (ii) is:

x(t) = Xocos(wt) + Xo

For part d, we can find the time t1 and position x1 at which the particle comes to rest by setting x'(t) = 0 and solving for t and x. Using the solution from

 

[kerimek]

Hello, thank you for sharing your thoughts on this problem. You are correct in your understanding of the absence of friction in simple harmonic motion (SHM). In the absence of friction, the total force on the object is equal to -2kx, which is equal to the mass times the square of the angular frequency (w) times the displacement (x). This can be written as -mw^2x.

For part (b), you have correctly identified the initial motion of the object when considering the effects of friction. When 2kXo is greater than the product of the coefficient of friction (\mu) and the weight of the object (mg), the object will oscillate with a lower frequency and its amplitude will gradually decrease. This is due to the frictional force opposing the motion of the object, causing it to lose energy and decrease its amplitude.

When 2kXo is less than or equal to the product of \mu and mg, the object will remain stationary. This is because the frictional force will be strong enough to keep the object in place, preventing any motion.

For part (c), you are on the right track in setting up the equation for the displacement of the object in the presence of friction. As you have correctly stated, the total force on the object will now be equal to -2kx + \mu mg, where \mu mg represents the frictional force. This can be written as -mw^2x + \mu mg. To solve for the constants A, B, and C, you can use the initial conditions given in the problem. The object is released at time t=0 with a positive displacement Xo from equilibrium. This means that x(0) = Xo, and x'(0) = 0 (since the object is initially at rest). Using these initial conditions, you can solve for the constants A, B, and C in the general solution x(t) = Acos(wt) + Bsin(wt) + C.

Finally, for part (d), you can use the general solution x(t) to find the time t1 and position x1 at which the object next comes to rest. This will occur when the velocity of the object is equal to zero, so you can set x'(t) = 0 and solve for t. This will give you the time t1 at which the object comes to rest. To find the position x