PH1140: Oscillations and Waves Name: SOLUTIONS AT END Conference: Date: 31 March 2005 EXAM #1: D2006 INSTRUCTIONS: (1) The only reference material you may use is one 8½x11 crib sheet and a calculator. (2) Show your work clearly and give numerical results to at least three significant figures. (3) Attach the proper units to all quantities in your answers. (4) Use the back of the preceding page for scratch paper. PROBLEMS: 1. (20) 2. (40) 3. (20) 4. (20) Total: (100)
1. Given: A cos( ωt + φ ) = 3 cos( ωt + π/3 ) + 2 sin( ωt π/2 ) + 1.5 cos( ωt π/5 ) a) Determine the amplitude A and the initial phase shift φ of the sum. b) If the sum above is now multiplied by 4 sin( ωt ), determine the frequency and amplitude of the resulting oscillation.
2. A particle moves sinusoidally according to the function: x(t) = 2.75 sin( 2.5π t π/3 ) (measured in centimeters). a) Find the period T, angular frequency ω, and the initial phase angle of x(t) b) Find the maximum value of the velocity and acceleration. c) What is the smallest positive value of t for which the acceleration is zero? d) If this x(t) is as a result of a 3 gram mass on a spring (ideal), find the spring constant k.
3. Two vibrations are given by: 4 sin( 100π t + 51 ) and A cos( ωt + φ ) which produce beats when added with a beat period of 0.75 seconds. The maximum value of the slowly varying amplitude is 10.5: a) Determine a value for A. and what can you conclude about ω? b) Find φ if the slowly varying amplitude takes on its minimum value at t = 2 s.
4. The x - y coordinates of a particle traveling in real space is given by: x(t) = 1.5 cos( ωt + φ ) and y(t) = 1.5 cos( ωt ± π/2 ) (in meters) where φ is positive. a) If the path is a circle moving in the clockwise direction determine which sign in y(t) to use and calculate φ. Explain your reasoning b) If the frequencies are now allowed to be different and the two initial phase shifts are zero ( φ x = φ y = 0 ), sketch below the pattern that would emerge if ω x = 3ω y. The amplitudes remain the same. Carefully note any special features. 2.0 Y 1.5 1.0 0.5 0.0-2.0-1.5-1.0-0.5 0.0 0.5 1.0 1.5 2.0-0.5 X -1.0-1.5-2.0
PH 1140 Exam #1 Solutions: 1) a) Write each term on the right-hand-side separately and convert to the physical standard form: u 1 (t) = 3 cos( ωt + π/3 ) = 3 cos( ωt + π/3 + π ) = 3 cos( ωt + 4π/3 ) u 2 (t) = 2 sin( ωt π/2 ) = 2 cos( ωt π/2 π/2 ) = 2 cos( ωt π ) u 3 (t) = 1.5 cos( ωt π/5 ) - (already in proper form) = 1.5 cos( ωt π/5 ) Since each oscillation has the same frequency ( Case I ), we evaluate at t = 0 s and write down each in complex, z = a + bj, notation: z 1 = 1.5 2.59807 j z 2 = 2 0 j z 3 = +1.21353 0.88168 j z T = 2.28647 3.47975 j So, the resulting amplitude and initial phase angle (since it has been evaluated at t = 0 s) is: A = (a T 2 + b T 2 ) ½ = 4.16372 = 4.164 φ = tan 1 ( b T / a T ) = 0.98946 + π ( should be in quadrant III ) = 4.1311 rad Finally, u T = 4.164 cos( ωt + 4.131 ) b) Converting to the physical standard form: 4 sin(ωt) = 4 cos( ωt π/2 ) then multiplying with the result from part a) gives: 16.6549 cos( ωt + 4.1311 ) cos( ωt π/2 ) = A B cos( θ a ) cos( θ b ) Using Euler s Relation, exp( jθ ) = cos(θ) + j sin(θ), we have 2 cos(θ) = exp( jθ ) + exp( jθ ) Substituting into the product above gives A B cos(θ a ) cos(θ b ) = (R/4) [exp(jθ a ) + exp( jθ a )] [exp(jθ b ) + exp( jθ b )] Carrying out the product and collecting terms gives: = (A B/4) [exp(j(θ a +θ b )) + exp( j(θ a +θ b )) + exp(j(θ a θ b )) + exp( j(θ a θ b ))] = (A B/2) [cos(θ a +θ b ) + cos(θ a θ b )] Finally, = 8.327 cos( 2ωt + 2.5603 ) + 8.327 cos( 5.7019 )
2) The position function should first be converted to the physical standard form: x(t) = 2.75 sin( 2.5π t π/3 ) = 2.75 cos( 2.5π t π/3 + π π/2 ) a) angular frequency = ω = 2.5π rad/s = 7.854 rad/s. period = T = 1 / f = 2π / ω = 2/3 s = 0.80 s. initial phase angle φ = π/3 + π/2 = π/6 = 0.5236 rad. ( or 30 o ) b) v max = v o = ωa = 6.875π cm/s = 21.598 cm/s a max = a o = ω 2 A = 17.1875π 2 cm/s 2 = 169.634 cm/s 2 c) the acceleration as a function of time is given by: a(t) = a o cos( 2.5π t + π/6 + π ) = a o cos( 2.5π t + 7π/6 ) which equals zero when the total phase angle is θ(t) = π/2 = 2.5π t + 7π/6. Solving gives t = 0.26666 s which is negative (not what was asked for) The acceleration is zero twice every period, so the first positive time is this t plus one half a period, T, or 0.26666 + (0.8) / 2 = 0.1333 s. d) mass = 3 gram = 0.003 Kg For the ideal mass-spring system ω 2 = k/m So, k = mω 2 = 0.01875π 2 Kg/s 2 = 0.1851 Kg/s 2 (or N/m)
3) Write each oscillation in the physical standard form: u 1 (t) = 4 sin( 100π t + 51 ) = 4 cos( 100π t + 51 + π/2 ) u2(t) = A cos( ωt + φ ) where the beat period T b = 0.75 s means f b = 1.333 Hz or ω b = 2.6666π rad/s and the maximum value of the sum is 10.5 a) when the two are in phase ( θ = 0 rad) they will produce a maximum which is the sum of the individual amplitudes. so, 10.5 = 4 + A giving A = 6.5 the second oscillation can have a value of the first frequency plus or minus the beat frequency. ω = 102.666π rad/s or ω = 97.3333π rad/s more information is needed to distinguish which really occurs. b) for a minimum net amplitude at t = 2 s, the two oscillations must be out of phase by θ = ±π. θ = ( ωt + φ ) ( 100π t + 51 + π/2 ) = π which simplifies to (we choose +π because the phase angle progresses forward when time elapses) ( ω 100π) (2 sec) + φ = π/2 51 although ω is not known (see part a), the difference is known ω 100π = ω b = 2.6666π rad/s φ = ( π/2 51 ) 5.3333π = 66.1844 rad (we subtract off whole multiples of 2π, leaving only the fraction of 2π) φ = 10.53357(2π) = 0.53357(2π) = 1.06713π = 3.3525 rad
4) a) The conditions on the x and y functions for a circular path are: same amplitudes same frequencies a phase difference of θ = θ x θ y = ±π/2 The first two are satisfied by inspection. The third requires that φ = π, 0, or +π. However, since it is stated that φ is positive, there are only two choices: φ = 0 or +π For clockwise path using φ = 0 at t = 0, the x-coordinate is x(0) = 2, then the y-coordinate starts at zero but must take on increasingly negative values. This means that y(t) acts like sin(ωt), and so we must use φ = 0 in x(t) and +π/2 in y(t) For clockwise path using φ = π at t = 0, the x-coordinate is x(0) = 2, then the y-coordinate starts at zero but must take on increasingly positive values. This means that y(t) acts like +sin(ωt), and so we must use φ = π in x(t) and π/2 in y(t) (you may check this by plugging in a few values for ωt and trace a plot). b) Using either the graphical method or a simple table of values gives: