# sample exam #1

```PH1140: Oscillations and Waves
Name:______Solutions___________
Conference: ______________
Date: __1 April 2005_______
EXAM #1:
D2005
INSTRUCTIONS:
(1) The only reference material you may use is one 8&frac12;x11 crib sheet and a calculator.
(2) Show your work clearly and give numerical results to at least three significant figures.
(4) Use the back of the preceding page for scratch paper.
PROBLEMS:
1. ________ (40)
2. ________ (20)
3. ________ (20)
4. ________ (20)
Total: _________ (100)
1. A particle moves sinusoidally according to the function:
x(t) = −1.75&middot;sin( 3π&middot;t − π/6 )
(measured in meters).
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 2 gram mass on a spring (ideal), find the spring
constant k.
2. Given:
A&middot;cos( ωt + φ ) = −4&middot;cos( ωt + π/4 ) + 4&middot;sin( ωt – π/3 ) + 2&middot;cos( ωt – π/6 )
a) Determine the amplitude A and the initial phase shift φ of the sum.
b) If the sum above is now multiplied by 2&middot;sin( ωt ), determine the frequency and
amplitude of the resulting oscillation.
3. Two vibrations are given by:
−5&middot;sin( 1100π&middot;t + 250 )
and
A&middot;cos( ωt + φ )
which produce beats when added with a beat period of 0.5 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) = 2&middot;cos( ωt + φ )
and
y(t) = 2&middot;cos( ωt &plusmn; 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 ( φx = φy = 0 ), sketch below the pattern that would emerge if ωx = 2ωy.
The amplitudes remain the same. Carefully note any special features.
Y
2.0
1.5
1.0
0.5
-2.0
-1.5
-1.0
0.0
-0.5 0.0
-0.5
-1.0
-1.5
-2.0
0.5
1.0
1.5
2.0
X
PH 1140 Exam #1 Solutions:
1)
The position function should first be converted to the physical standard form:
x(t) = −1.75&middot;sin( 3π&middot;t − π/6 ) = 1.75&middot;cos( 3π&middot;t − π/6 + π − π/2 )
a)
period = T = 1 / f = 2π / ω = 2/3 s = 0.6667 s.
initial phase angle φ = − π/6 + π − π/2 = π/3 = 1.0472 rad. ( or 60o )
b)
vmax = vo = ωA = 5.25π m/s = 16.4934 m/s
amax = ao = ω2A = 15.75π2 m/s2 = 155.447 m/s2
c)
the acceleration as a function of time is given by:
a(t) = ao&middot;cos( 3π&middot;t + π/3 + π ) = ao&middot;cos( 3π&middot;t + 4π/3 )
which equals zero when the total phase angle is θ(t) = π/2 = 3π&middot;t + 4π/3.
Solving gives t = −0.2777 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
−0.2777 + (0.6667) / 2 = 1/18 s = 0.0557 s.
d)
mass = 2 gram = 0.002 Kg
For the ideal mass-spring system ω2 = k/m
So, k = mω2 = 0.018π2 Kg/s2 = 0.1777 Kg/s2.
2)
a)
Write each term on the right-hand-side separately and convert to the
physical standard form:
u1(t) = −4&middot;cos( ωt + π/4 )
u2(t) = 4&middot;sin( ωt – π/3 )
u3(t) = 2&middot;cos( ωt – π/6 )
= 4&middot;cos( ωt + π/4 + π ) = 4&middot;cos( ωt + 5π/4 )
= 4&middot;cos( ωt – π/3 − π/2 ) = 4&middot;cos( ωt – 5π/6 )
(already in proper form)= 2&middot;cos( ωt – π/6 )
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:
z1 =
z2 =
z3 =
−2.8284
−3.4641
+1.7321
−2.8284 j
−2 j
−1 j
zT =
−4.5604
−5.8284 j
So, the resulting amplitude and initial phase angle (since it has been
evaluated at t = 0 s) is:
A = (aT2 + bT2) &frac12; = 7.4005 = 7.4
φ = tan−1 ( bT / aT ) = 0.9069 + π ( should be in quadrant III ) = 4.0484 rad
Finally,
b)
uT = 7.4&middot;cos( ωt + 4.05 )
Converting to the physical standard form: 2&middot;sin(ωt) = 2&middot;cos( ωt – π/2 )
then multiplying with the result from part a) gives:
14.8&middot;cos( ωt + 4.05 )&middot;cos( ωt – π/2 ) = A&middot;B&middot;cos( θa )&middot;cos( θb )
Using Euler’s Relation, exp( jθ ) = cos(θ) + j sin(θ), we have
2&middot;cos(θ) = exp( jθ ) + exp( −jθ )
Substituting into the product above gives
A&middot;B&middot;cos(θa)&middot;cos(θb) = (R/4)&middot;[exp(jθa) + exp(−jθa)]&middot;[exp(jθb) + exp(−jθb)]
Carrying out the product and collecting terms gives:
= (A&middot;B/4)&middot;[exp(j(θa+θb)) + exp(−j(θa+θb)) + exp(j(θa−θb)) + exp(−j(θa−θb))]
= (A&middot;B/2)&middot;[cos(θa+θb) + cos(θa−θb)]
Finally,
= 7.4&middot;cos( 2ωt + 4.05 ) + 7.4&middot;cos( 4.05 − π/2 )
3)
Write each oscillation in the physical standard form:
u1(t) = −5&middot;sin( 1100π&middot;t + 250 ) = 5&middot;cos( 1100π&middot;t + 250 + π/2 )
u2(t) = A&middot;cos( ωt + φ )
where the beat period Tb = 0.5 s means fb = 2 Hz or ωb = 4π 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 = 5 + A
and
A = 5.5
the second oscillation can have a value of the first frequency plus or minus
the beat frequency.
or
b)
for a minimum net amplitude at t = 2 s, the two oscillations must be out of
phase by ∆θ = &plusmn;π .
∆θ = ( ωt + φ ) − ( 1100π&middot;t + 250 + π/2 ) = π
(we choose +π because
the phase angle
progresses forward
when time elapses)
which simplifies to
( ω − 1100π)&middot;(2 sec) + φ = π/2 − 250
although ω is not known (see part a), the difference is known
| ω − 1100π | = ωb = 4π rad/s
φ = ( π/2 − 250 ) − 8π = −273.5619 rad
(we subtract off whole multiples of 2π, leaving only the fraction of 2π)
φ = −43.5387(2π) = −1.0775π = −3.3850 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 = θx − θy = &plusmn;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:
2.0
Y
1.5
1.0
0.5
-2.0
-1.5
-1.0
0.0
-0.5 0.0
-0.5
-1.0
-1.5
-2.0
0.5
1.0
1.5
2.0
X
```