rowellVT

__**Second week : (05th of September, 2011)**__
 * 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.

//Steps// 1. Find the mass of one log using the equation mass = density*Volume (Volume = pi*r^2*L) 2. Multiply mass by 6 to find the mass of the raft 3. To find the weight of the raft use the formula Weight = (density)(g)(Volume), using the Volume found in step 2, the density of logs, and 9.81 m/s^2 for g 4. Now find the buoyant force using a similar formula Fb = (density)(g)(Volume), using the Volume found in step 2, the density of salt water, and 9.81 m/s^2 for g 5. Find the net force by subtracting the Weight from the buoyant force (Fb). This is the amount of weight the raft can hold 6. Divide by 200N (the weight of one child) and round down to the nearest whole number (because you can't have partial children) 7. Done


 * 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?

//Steps// 1. The idea behind this problem is that the buoyant force = the weight of the cube 2. Start by using the equation (density of wood)(g)(Volume of cube) = (density of displaced liquid)(g)(height of liquid)(Area of cube face in contact with liquid) 3. On the liquid side of the equation, the density is 1240kg/m^3, Area is just s^2 and the height is what we're solving for 4. On the cube side, the density is 550kg/m^3 and the Volume is s^3 5. Solve for height 6. Done


 * Professor's Note : Sam, good job, you are doing everything right here and wish you did the same in the Quiz. Keep up the good work **

**Third week (12th September, 2011) **

**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?

//Steps// 1. Find a formula that gives you what you want from what you have [vmax = xmax*omega & omega = sqrt(k/m), where vmax = velocity of car at impact, xmax = compression of bumper] 2. Plug found value of omega from second formula into the first formula 3. Done

**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.

//Steps// 1. Convert amplitude into meters 2. Use formula F = -kx to find F 3. m = F/g 4. Done

ii. Find the period of the motion.

//Steps// 1. T = 1/f and f = omega/2pi and omega = sqrt(k/m) 2. Combine all formulas and solve for T 3. Done

iii.Calculate the maximum acceleration of the block.

//Steps// 1. amax = omega^2 * xmax and omega = sqrt(k/m) 2. Combine all formulas and solve for amax 3. Done

**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.

// Steps // 1. vmax = xmax * omega and omega = sqrt(k/m) 2. Combine these two equations and solve for vmax 3. Done

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

//Steps// 1. Use formula x(t) = xmax*cos(omega*t) and solve for t 2. Plug t into the formula v(t) = -xmax*omega*sin(omega*t) and solve for v(t) 3. Done

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

// Steps // 1. Take the value of omega and x(t) from part ii. and plug it into the formula a(t) = -(omega^2)*x(t) 2. Done

__ **Fourth week (19th September, 2011) ** __

01. A transverse wave of frequency 35 Hz propagates down a string. Two points 5 cm apart are out of phase by π/3

i. What is the wave length of the wave? (Hint : one full wavelength is 2π apart !)

//Steps// 1. Set up proportion: (distance between points)/(pi/3) = (wavelength)/(2pi) 2. Solve for wavelength 3. Done

ii. At a given point, what is the phase difference (as a multiple of π ) between two displacements for times 5 ms apart?

//Steps// 1. Find the period using T = 1/f 2. Remember that one oscillation covers a phase of 0 to 2pi 3. Set up a proportion: (phase difference/0.05s)=(2pi/period) 4. Solve for phase difference 5. Divide by pi to get your answer in terms of pi radians 6. Done

02. 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) )

//Steps// 1. Take the derivative of the y(x,t) function to get v(x,t) which equals 4.5 2. Solve for x 3. Plug the value you get for x into the original function 4. Solve for y(x,t) 5.Done

// **Professor's Note** : First one is perfectly done and the second is correct and I do hope you did the math to the end, it is little bit tricky as it involves an integration //** !! **

**Fifth week (26th September, 2011) --- please consider this as a review for your exam too **

1. A light string of mass per unit length 8 g/m has its ends tied to two walls separated by a distance equal to three fourths the length L of the string. A mass m is suspended from the center of the string, applying a tension in the string. The acceleration of gravity is 9.8 m/s². What size mass should be suspended from the string to produce a wave speed of 60 m/s?

//Steps// 1. The system is not moving, there for Sum of Forces = 0 2. The tension in the string, T, is in the positive y direction, while the weight, mg, is in the negative y direction. Therefore Sum of Forces = 2Tsin(theta)-mg where theta is the angle the stretched string makes with the horizontal and can be found using the law of cosines: arccos((3L/8)/(L/2)) = theta. 3. Solve for T -> T=mg/(2sin(theta)) 4. Plug T into velocity formula: v = sqrt((mg/(2sin(theta)))/mu) 5. Solve above equation for mass -> m = 2v^2(mu)sin(theta)/g 6. Done

<span style="display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">2. A simple pendulum consists of a ball of mass M = 5 kg hanging from a uniform string of mass m = 0.6 g and unknown length L. The pendulum is located in gravitational field g = 9.8 m/s² and has period T = 2 s. <span style="color: #0000cc; display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. (For <span style="display: block; font-family: 'times new roman',times,serif; font-size: 14px; text-align: left;">simplicity, neglect the string’s weight compared to the ball’s weight. Likewise, neglect the ball’s radius compared to the string’s length). Answer in units of m/s. //Steps// 1. Use the formula T=2pi*sqrt(L/g) to find L 2. Find mu by using the formula: mu = m/L 3. Find the tension, T, in the string by noticing: T=Mg 4. Plug all the info calculated into the formula: v = sqrt(T/mu) 5. Done

Week seven ( 10th October, 2011)

01. The “red shift” of radiation from a distant galaxy consists of the light known to have a wavelength of 434 nm when observed in the laboratory, appearing to have a wavelength of 462 nm. What is the speed of galaxy in the line of sight relative to the Earth? The speed of light is 2.99792 × 10^8 m/s.

//Steps// 1. Convert wavelength of both lights into SI units, then find their frequencies using: frequency using v=f*lambda, where v = c = 2.99792E8 m/s 2. Use the Doppler effect formula for moving source, stationary observer since the universe is expanding. It is: f_o = f_s*[1/(1+(v_s/v))] 3. Solve previous formula for v_s 4. Done

2. The sound of a man shouting at the top of his lungs from a rather large distance away from your ear has loudness of only 20 decibels. What would be the decibel level of four men shouting at the top of their (equally powerful) lungs from the same distance away from you ear? Assume that there is no interference from superposed waves.

//Steps// 1. Recall the formula for total sound power level: Beta = 10log(I/I_o) 2. If you had N times the people shouting, I is now N times larger 3. New formula: Beta_total = 10log(N*I/I_o) 4. Simplify: Beta = 10log(I/I_o) + 10log(N) using properties of logarithms 5. Simplify further: Beta_total = Beta_single +10log(N) 6. Plug in value for N from problem 7. Done

Professor's Note : Very good answers !

Week eight ( 18 October, 2011)

Define polarization

The phenomenon in which waves of light or other radiation are restricted in direction of vibration.

1. Initially unpolarized light is sent into a system of three polarizing sheets whose polarizing directions make angles of Ø1 = Ø2 = Ø3 = 50⁰ with the directions of the y axis. What percentage of the initial intensity is transmitted by the system? (hint: be careful with the angles)

//Steps// 1. The formula for the intensity of unpolarized light as it passes through a polarizing sheet is: I_1 = (1/2) I_0, i.e. the light coming out the sheet has half the intensity as the unpolarized light going in. 2. Also, for polarized light going through a polarizing sheet, we use the formula: I_1 = I_0 (cos (theta))^2, where theta is the angle between the polarization of the incident light and the polarization axis of the polarizing sheet 3. So, the progression would be as follows: I_1 = 1/2 I_0 -> I_2 = I_1 (cos (0))^2 or 1/2(I_0)(cos(0))^2-> I_3 = I_2 (cos (0))^2 or 1/2(I_0)(cos(0))^2*(cos(0))^2 4. Solve for I_3 in terms of a percent of I_0 5. Done

__** Week nine ( 24th October, 2011) **__ 1. A fruit fly of height H sits in front of a lens (say lens 1) on the central axis. The lens forms an image of the fly at a distance d = 20 cm from the fly ; the image has the same orientation as the fly and a height = 2 H.

a) What are the focal length of the lens and the object distance of the fly?

//Steps// 1. Recall that m = h'/h AND m = -i/p. Therefore, h'/h = -i/p 2. Solve the previous formula for p to find the focal length 3. Plug p and i into the formula 1/f = 1/p +1/i 4. Solve for f 5. Done

Fly then leaves lens 1 and sits in front of a second lens (say lens 2), which also forms an image at d = 20 cm from the fly and has the same orientationas of the fly, but now the image height is 0.5 H

b) What are the focal length of the second lens and the new object distance?

//Steps// 1. Solved in the same manner as part a. 2. Done

2. When the rectangular metal tank in the figure, is filled to the top with an unknown liquid, Observer O with eyes level with the top of the tank, can just see corner E. Aray that refracts toward O at the top surface of the liquid is shown. If D = 85.0 cm and L = 1.1 m, what is the index of refraction of the liquid?

//Steps// 1. Find the angle at which the light is hitting the liquid-air interface using sin(theta) = (opposite side)/(hypotenuse). We can observe that the angle of refraction in 90 degrees 2. Use Snell's Law (n1sin(angle of incidence) = n2sin(angle of refraction)) to find n1 3. Done