AdamPaul

Adam Paul Physics II PHY 113 Fall 2011

11:23 pm --- don't see an update ! -- prof


 * Week Four:**

i. What is the wave length of the wave? (Hint : one full wavelength is 2π apart !) ii. At a given point, what is the phase difference (as a multiple of π ) between two ‘displacements for times 5 ms apart?
 * 1.** A transverse wave of frequency 35 Hz propagates down a string. Two points 5 cm apart are out of phase by π/3

I. Well, if the total phase/wavelength is 2π, then the two points make up one sixth of the total phase. That means that the wavelength is .3m.

II. Period is the inverse of frequency, so T=1/35 or 28.57 ms. Which is the whole wavelength 2π. So we can make a proportion x/5ms=2π/28.57 and solve for x and thats your phase difference.
 * Here you have all correct steps !! **
 * 2.** At t = 0, a transverse wave pulse in a wire is described by the function y (x, t= 0) = 6/( x²+ 3), where x and y are in meters. The pulse is traveling in the positive x direction with a speed of 4.5 m/s. What is y at time = t? (Hint : find y(x,t) )

Well all we really have to do is make the equation work for only values of t, so we just have to find the relationship between x and t and substitute all x-values for equivalent t-values. If V=4.5 m/s, then for every time=t seconds, x will be 4.5t. So we plug that in for x in the original equation and boom goes the dynamite:

y(t) = 6 / (20.25*t^2 + 3) This is not the right answer !! -- professor

**1.**  An automobile having a mass of 1000 kg is driven into a brick wall in a safety test. The bumper behaves like a spring of constant 5 × 10^6 N/m and compresses 3.16 cm as the car is brought to rest. What was the speed of the car before impact, assuming no energy is lost during impact with the wall?
 * Week Three:**

Use the equation 1/2*m*V^2 = 1/2*k*x^2. Solve for Vmax... V= √((k*x^2)/m) and plug in your values and calculate.

* 2. A block of unknown mass is attached to a spring of spring constant 6 .5 N/m and undergoes simple harmonic motion with an amplitude of 10 cm. When the mass is halfway between its equilibrium position and the endpoint, its speed is measured to be v = 30 cm/s. i. Calculate the mass of the block.

Use the equation 1/2*m*V^2 = 1/2*k*x^2. Solve for Mass, m.... m = (k*x^2) / V^2 and plug in your values and calculate. Don't forget to convert amplitude and velocity into meters and meters per second!!!!

ii. Find the period of the motion.

Use the equation omega = √(k/m) and use m from part 1. Then plug omega into the equation... Frequency, f = omega / 2π. Then invert the frequency to get Period, T in seconds.

iii.Calculate the maximum acceleration of the block.

Max acceleration occurs at max displacement from equilibrium, which is also where max force occurs. So if we take the objects mass from part one, we can use it in F=ma and F=-kx... SOOOO ma=-kx when x is 10cm. Solve for max acceleration.

3. A 0.4 kg block attached to a spring of force constant 12 N/m oscillates with an amplitude of 8 cm. i. Find the maximum speed of the block.

Convert amplitude into meters and use the equation 1/2*m*V^2 = 1/2*k*x^2. Solve for Vmax... V= √((k*x^2)/m) and plug in your values and calculate.

ii. Find the speed of the block when it is 4 cm from the equilibrium position.

At any point in the block's motion, the total energy (kinetic and potential) is the same. Soooooo total energy = 1/2*k*x^2 which also equals KE plus PE soooooo 1/2*k*x^2 = 1/2*m*V^2 + 1/2*k*x^2 Use the answer from part one to substitute for the left side of the equation and .04m for the amplitude on the right side. Then just solve for V.

iii. Find its acceleration at 4 cm from the equilibrium position.

Use F=ma and F=-kx. Soooo ma = -kx just plug in values and solve for a.


 * Week Two:**


 * 1.** A log-raft was made by lashing six logs together, where each log has a diameter 0.25 m and length 1.90 m. How many children can safely afloat on the raft in sea water, if the average weight of a child is 200 N? Density of the logs is 800 kg/m3 and sea water is 1024 kg/m3.

Step 1. We need to figure out the maximum buoyant force provided by six logs, so we can counter it with as many children as possible. Max buoyant force occurs when all the volume of an object (in this case the entire raft, all six logs) is just barely submerged beneath the surface of the water. So we'll calculate the total volume of the raft:

Six logs, each a cylinder, (A=pi*r^2) (Don't forget to halve the diameter) and then multiply by six.

Step 2. The buoyant force is equal to the weight of the displaced water by the raft, so if we assume that all of the raft's volume is displacing water, we can multiply the raft volume and density of salt water to find the displaced water's mass. Then multiply by g to get the weight, which is also the max buoyant force.

Step 3. The raft will then support an equal number of negative Newton's to the max buoyant force (in positive Newton's). The forces in the negative y direction are the combined weights of the children and the raft. Remember Density = Mass/ Volume, so volume of the raft (Vt) times Density (p) equals the rafts mass.

Step 4. Subtract the raft's mass from the maximum buoyant force and you have the remaining buoyant force to lift children.


 * 2.** A cube of wood whose edge is 0.015 m is floating in a liquid (in a glass beaker) with one of its faces parallel to the liquid surface. The density of wood is 550 kg /m³, that of liquid is 1240 kg/m³. How far below the liquid surface is the bottom face of the cube?

Step 1. You can make a proportion between the height of the cube that is submerged at equilibrium and at max buoyant force, so you'll need the weight of the cube and the max buoyant force.

Step 2. Weight of the cube is it's mass times g. It's mass can be expressed as density times volume. So it's weight is 9.8*550*(.015^3). (Volume for a cube is s^3) This weight is equivalent to the necessary buoyant force to keep the block floating at equilibrium.

Step 3. Next, we need the max buoyant force so we'll take the cube to be completely submerged this time and use V*p*g and use the cube's whole volume, and the density of the liquid.

Step 4. Now we can set up our proportion.

.(Weight of the block) (Height of cube that is submerged at equilibrium) <-- This is the value we're looking for. -- = -- .(Max Buoyant Force) (Total Height of Cube)


 * //Professor's Note : As always doing a great job, All equations, concepts are explained very well. Note : try to submit the answer by 9 pm//. **