Assignment 2 -- Solutions

Exercise 1 page 68.  Both oscillators have the same period between successive cycles, namely 1.2 seconds.  Mass 1 reaches its lowest position 0.9 seconds before mass 2.  So mass 1 leads mass 2 by 0.9/1.2 times 360 degress = 270 degrees or 3π/2 radians.

Exercise 2 page 69.  1/6 of a cycle is 1/6 x 360 degrees = 60 degrees or π/3 radians.

Exercise 4 page 69.  Superposition of two identical frequencies.  Choose x between 0. and 1. to set the relative phase, then execute the cell.  One can see that the superposition is always a sine wave of frequency 440 cycles/second or 440 Hz.

x=.25;
Plot[{Sin[2 Pi 440 t], Sin[2 Pi 440t + x 2 Pi], Sin[2 Pi 440 t]+Sin[2 Pi 440t + x 2 Pi]},{t,0.,0.01}]

Exercise 5 page 69.  We add two SHMs  of amplitude 0.2 mm each with frequencies separated by a perfect fifth, i.e. (Subscript[f, 2]) = 3/2 (Subscript[f, 1]).  We will let (Subscript[f, 1]) = 440 Hz.  The initial phase difference is set at π radians.  The plot below show that the superposition is a complex motion which is periodic with period twice that of the (Subscript[f, 1]) SHM.  This corresponds to a frequency of 220 Hz.

Plot[{0.2Sin[2 Pi 440 t], 0.2Sin[2 Pi 3/2 440t + Pi], 0.2Sin[2 Pi 440 t]+0.2Sin[2 Pi 3/2 440t + Pi]},{t,0.,0.01}]

Since the only thing that matters here is the ratio of the frequencies, we can also make a plot in terms of a parametric argument α and 3/2α as α varies from 0 to say 5π radians.  The most interesting thing to see is that the superposition is a complex wave form with a frequency which is 1/2 (Subscript[f, 1]).  In the parametric plot below this shows up as the fact that the superposition repeats itself every 4π radians in the parameter α.

Plot[{0.2Sin[α],0.2Sin[3/2α+Pi],0.2Sin[α]+0.2Sin[3/2α+Pi]},{α,0,5Pi}]

Exercise 10 on page 44 of Rigden.  A source of sound has an intensity of 0.5 W/(m^2) at a distance of 20 m from the source. (a)  What is the intensity at 10 m.  Since the sound intensity fall off as 1/ (distance^2) we calculate (see below) Intensity = 2.0 w/(m^2) at a distance of 10 m.  Inother words, since the point at which we wish to calculate the intensity is half the distance from where we know it, its value is 4 times as great.

Intensity= 0.5 W/(m^2) 20.^2/10.^2

(b) What is the intensity at 30 m?  Again we use the ratio of the square of the distances (see below).  Express your answer to no more than three decimal places, i.e., Intensity = 0.222 W/(m^2).

Intensity= 0.5 W/(m^2) 20.^2/30.^2

Exercise 11 on page 44.  What is the sound-intensity level (SL) of a sound with an intensity = (10^(-5)) W/(m^2)?  We use SL= 10 log (I / (Subscript[I, 0])) where (Subscript[I, 0]) = (10^(-12)) W/(m^2) to get SL in decibels or db (see below). The answer is SL = 70 db.

Intensity = (10^(-5)) W/(m^2);
SL=10 Log10[Intensity/((10^(-12)) W/(m^2))] db

Exercise 2 on page 56.  What is the SL of a sound with intensity = 6 X (10^(-5)) W/(m^2)?  We do the same calculation as in the previous exercise with only the intensity changed (see below).  Rounding the answer gives SL = 77.8 db.  The point the book is making in the lead up to the question is that you can get the answer by adding (Subscript[log, 10])(6) to (Subscript[log, 10]) ((10^7)) and then multiplying the result by ten.

Intensity = 6.0 (10^(-5)) W/(m^2);
SL=10 Log10[Intensity/((10^(-12)) W/(m^2))] db