SimonGangon


 * 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?

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. Determine the speed of a transverse wave in the string when the pendulum is stationary and hangs vertically. (For 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.

Note from your professor : Simon, try your best to not to give a final numerical answer but to provide the steps to reach the final answer. Over all great job so far !! **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 !)

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

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

**Simple Harmonic motion **

**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?

K=5*10^6 N/m X=3.16 cm M=1000 kg
 * Given: **

energy of compression = (.5)*K*(X^2) kinetic energy = (.5)*M*(V^2) Set the energy of compression = kinetic energy and solve for V.
 * Setup: **

(.5)*(5*10^6)*.0316 = (.5)*1000*(V^2) (5*10^6)*.0316 = 1000*(V^2) ((5*10^6)*.0316)/1000 = V^2 square root both sides V = 12.569 m/s --- <span style="font-family: 'times new roman',times,serif; font-size: 14px;">**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. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">i. Calculate the mass of the block.
 * Answer:**

K = 6.5 N/m X = 10 cm V = 30 cm/s
 * Given:**

energy of compression = (.5)*K*(X^2) Find the maximum (at 10 cm) and the halfway (at 5 cm) Find the difference (this is equal to kinetic energy) and set it equal to (.5)*M*(V^2)
 * Setup:**

((.5)*6.5*(.1^2)) - ((.5)*6.5*(.05^)) = .024375 .024375 = (.5)*M*(.3^2) M = .5417 Kg
 * Answer:**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">ii. Find the period of the motion.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">Plug values into the formula for Period
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">Setup: **

T = 2*pi*sqrt(M/K) T = 2*pi*sqrt(.5417/6.5) T = 1.814 s
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">Answer: **

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">iii.Calculate the maximum acceleration of the block.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">Set KX= Ma and solve for a
 * <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">Setup: **

6.5*.1 = .5417*a a = 1.1999 m/s^2
 * Answer:**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">**3.** A 0.4 kg block attached to a spring of force constant 12 N/m oscillates with an amplitude of 8 cm. <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">i. Find the maximum speed of the block.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">**Given:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">M = .4 Kg K = 12 N/m X = 8 cm

w = sqrt(K/X) w = 12.247
 * Setup:**

12.247*.08 = .98 m/s
 * Answer:**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">ii. Find the speed of the block when it is 4 cm from the equilibrium position.

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">**Answer:** <span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">12.247*sqrt((.08^2)-(.04^2)) = .848 m/s

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">iii. Find its acceleration at 4 cm from the equilibrium position.

acceleration = - a w^2 sin(w*t) = -(w^2)*x
 * Setup:**

<span style="font-family: 'Times New Roman',Times,serif; font-size: 14px;">a = .6 m/s^2
 * Answer:**