A block of mass mA = 2 kg is attached to a spring with a...

A block of mass mA = 2 kg is attached to a spring with a force constant of k = 80 N/m and placed on a horizontal frictionless surface (gray block in the figure below). The block is made to oscillate such that the distance between points when the spring is most compressed and most relaxed is 0.8 meters (the range of motion). At time t = 0, the block is in its position furthest to the left, where the spring is maximally compressed. Answer the following questions: a) (3 points) Write the equation for the horizontal displacement from equilibrium x (t) in the form of x (t) = A cos (wt + phi). That is to say, determine the parameters A, w, phi in the expression. b) (7 points) At the instant when block mA is passing through the equilibrium position, a second block mB = 3 kg (green block in the figure) falls on top the first block. The two blocks stick together and the oscillating motion continues uninterrupted. What is the new amplitude A2 of oscillation of the mass of the two blocks combined? c) (4 points) How much time does the new two-block system need to move from the point where the spring is most stretched to the equilibrium point? Give your answer in seconds.

Transcript

00:01 To solve the shown problem and first we have to find an equation for the horizontal displacement of the block of the block m -a.

00:15 So let's do this.

00:17 In general, yeah, this equation has a form shown here.

00:21 It's a cosine of omega t plus phi.

00:28 So now let's look at let's look at the conditions of this oscillation so the block is made of to oscillate by the external force when the spring is most compressed yeah yeah that's 0 .8 meters is a total distance of oscillation so if you draw the x coordinate so let's say this is x this is y the distance of 0 .8 is a distance between the most compressed and most stretched spring.

01:27 Therefore, the middle is the position of the equilibrium.

01:37 So then the amplitude of the oscillation is for 0 .4 meters.

01:45 So here in this equation, a equals to 0 .4 meters.

01:52 4 meters.

01:57 At the initial moment of time the block is at its position further to the left where the spring is maximum pressed so it will be here.

02:14 So therefore at time equals to 0, a equals to minus 0 point x equals to minus 0 .4 so therefore if we have to change if it will be the angle which would make a sign negative therefore phi equals to negative because or cosine actually or pi it doesn't really matter because this cosine is negative 1 now we just need to and omega equals to square root of k over m which is square root of 18 newton meters newton meter over two kilograms let's calculate this number it equals to 6 .3 power by minus 5 oh therefore the overall equation of 0 .4 times of cosine of 6 .3c plus 5.

04:21 So that's the equation equation a.

04:25 Now let's move on and answer question b.

04:32 At the instant when block ma passing through the equilibrium point a second block is added and here we have to calculate the new amplitude of so when at equilibrium position total energy of the block equals to the sum of its kinetic energy plus some of the potential energy but the potential energy is zero therefore it equals maximum kinetic energy.

05:21 Then maximum kinetic energy of the block equals to its maximum potential energy of all the potential maximum potential energy of the deform spring so then we can further simplify it so m maximum velocity over two squared over two equals to k the a squared over two so therefore bm maximum velocity equals to square root of k over m multiplied by a or yeah square root of 80 newton meters newton over meter divided by mass of the block which is two kilograms right two kilograms multiplied by zero four meters let's calculate this number it equals to 2 .2 .53 meters per second or roughly 2 .5 meters per second that's the maximum velocity of the block actually no we'll just leave to 53 and we'll do the rounding for the new amplitude so that's the maximum that's the maximum energy before the second block is dropped onto the new one now let's the second block was added the total momentum was m1 va but after the second block is added the momentum becomes m1 plus m2 multiplied by some maximum velocity prime so the new maximum velocity of the system and now we will calculate this velocity so it equals to 2 kilograms times 2 .53 meters per second divided by the sum of two masses, which is 2 kilograms plus 3 kilograms...

Question

Please give Ace some feedback