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Questions: 1 2 3 4 5 6 7 8 9 10 11
| Physics 1101 | Oscillations (SHM) |
where distance is measured in metres and time in seconds.
(a) What is the amplitude, frequency, angular frequency, and period
of this motion?
(b) What is the equation of the velocity of this particle?
(c) What is the equation of the acceleration of this particle?
(d) What is the spring constant?
(e) At what next time t > 0, will the object be:
(a) First write the general expression on top of the given expression
| x = Acos(ωt + Δ ) |
| x = 4cos(1.33t + π/5) |
Immediately, we get A = 4 m, ω = 1.33 rad/s,
and Δ = π/5.
Since ω = 2π / T
= 2πf, we get T = 2π
/1.33 rad/s = 4.724 s, and f = ω
/ 2π = 0.2117 s-1.
(b) The velocity is given by the first derivative of position with respect to time
With the given values, we get
(c) The acceleration is given by the second of position with respect to time, or the first derivative of the velocity with respect to time,
With the given values, we get
(d) We have the relation that ω2 = K/m, so
(e) (i) & (ii) We know that at equilibrium x = 0. We also know that there are two places where this happens, one where is the velocity is positive and the object is moving to the right, and one where the velocity is negative and the object is moving to the left. So first let's set x = 0,
We can divide through by 4, and we get
Taking the inverse of both sides, the solution is
and thus,
Now cos-1(0) has many solutions, all the
angles in radians for which the cosine is zero.
Note however that your calculator will only give you one solution! You need to remember
that there are more solutions and how they are related.
This occurs for
angles θ = π/2,
θ = -π/2,
θ = 3π/2,
θ = -3π/2,
and so on. This is usually expressed
So our solutions for t are in the form
The first nonzero time when x = 0 occurs for n = +1,
The second nonzero time occurs when n = +2,
To tell which has the object moving to the right
and which to the left we examine the velocity
We see that the object is moving to the left, has
negative velocity, at t = t1 = 0.7086 s, and is moving
to the right at t = t2 = 3.0707 s.
An alternate way of solving this problem is to consult the unit circle. At t = 0, the unit circle looks like the diagram below. Note that the x components of the vector indicates that at t = 0, x(0) > 0, v(0) < 0, and a(0) < 0.
Now as we rotate counterclockwise with increasing time, there are two different positions on the unit circle where the position vector has no x component as shown below
The earliest time that the position of the object is at zero is solved from
which yields t1 = 0.7086 s. A glance at the left-hand graph indicates v < 0 here.
The next time the position is zero is found from
which yields t2 = 3.0707 s. A glance at the right-hand graph indicates v > 0 here.
All subsequent points are separated by time period T.
(iii) At maximum amplitude, x = +4, so we have
Dividing through by 4, we get
Taking the inverse of both sides, the solution is
and thus,
Now cos-1(1) has many solutions, all the
angles in radians for which the cosine is plus one. This occurs
for angles θ = 0,
θ = 2π,
θ = -2π,
θ = 4π,
θ = -4π,
and so on. This is usually expressed
So our solutions for t are in the form
The first nonzero time when x = +4 occurs for n = +1,
(iv) At minimum amplitude, x = -4, so we have
Dividing through by 4, we get
Taking the inverse of both sides, the solution is
and thus,
Now cos-1(-1) has many solutions, all
the angles in radians for which the cosine is negative one. This
occurs for angles
θ = π,
θ = -π,
θ = 3π,
θ = -3π,
and so on. This is usually expressed
So our solutions for t are in the form
The first nonzero time when x = -4 occurs for n = +1,
Our answers for (e) are thus
| (i) | t = 3.071 s, |
| (ii) | t = 0.709 s, |
| (iii) | t = 4.25 s, and |
| (iv) | t = 1.89 s. |
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Simple Harmonic Motion is independent of amplitude.
Our answers to Question #1 would not change.
The potential energy is spring potential energy and
is given by U = ½Kx2, so
The kinetic energy is given by K = ½mv2, so
The total energy is the sum of potential and kinetic
energies,
Examining the graph we see that the largest displacement
is 10 cm, so A = 0.10 m. We also see that the motion repeats every
0.2 seconds, so T = 0.2 s. Angular frequency is related to the
period by ω = 2π/T,
so ω = 10π.
Thus far we have
where Δ is still unknown. To find
Δ, we take a specific time and note the value
of x. For instance, x = -0.075 m at t = 0, so we have
This becomes cos(Δ) = -.75. Your caluclator will tell you that Δ = 138.6° or 0.77π. It is lying. There are infinitely many solutions to this problem. There are two solutions between -π and π, namely Δ1 = +0.77π and Δ2 = -0.77π. There are many other solutions Δ = ±0.77π + n2π, where n = 0, ±1, ±2, ±3, ? Any of these values is acceptable since the motion repeats itself every 2π. However, it is customary to keep Δ between -π and +π. To see which value is correct, notice from the graph that the velocity is positive at t = 0. We know that the velocity has the form v(t) = -ωAsin(ωt + Δ), so we can see which value of Δ gives the correct behaviour.
For Δ = +0.77π, v(0) = -ωAsin(0.77π) < 0, since sin(0.77π) is positive.
For Δ = -0.77π, v(0) = -ωAsin(-0.77π) > 0, since sin(-0.77π) is negative.
Or we can consider the unit circle for each case
Note that when Δ1 = +0.77π, the x component of the velocity is positive. When Δ2 = -0.77π, the x component of the velocity is negative.
Thus we conclude that
An examination of our relationships for SHM indicates that K = ω2m, E = ½KA2, vmax = ωA, and amax = ω2A, so
The equation for the velocity of an object undergoing SHM has the form v(t) = -vmaxsin(ωt+Δ), where vmax = ωA and ω = 2π/T. Examining the graph, we see that the period is T = 0.1 s, so ω = 20π s-1. Also the maximum velocity is 5 m/s. From this we determine that A = vmax/ω = (5 m/s)/( 20π s-1) = 1/(4π) m = 0.0796 m. Furthermore, the maximum acceleration is amax = ω2A = ωvmax = 100π m/s2.
The position equation for SHM is x(t) = Acos(ωt + Δ). We just found ω and A so all that is left to do is to find Δ. Again we look at the value of the graph at t = 0 which is v(0) = 2.40 m/s. So we have 2.40 = -5sin(Δ), or sin(Δ) = -0.48. There are two angles on the unit circle that satisfy this, Δ1 = -28.69° and Δ2 = -151.31°. We examine the unit circle for these possibilities
The graph above indicates that the acceleration, the slope of the line tangent to the velocity curve, at t = 0 is positive. Thus the correct phase angle is Δ2 = -151.31° = -2.64 rad. Our equation is x(t) = (0.0796 m)cos(20πt - 2.64).
The spring constant is given by ω2 = K/m, so K = (20π s-1)2(2.00 kg) = 7896 N/m.
The total energy is Etotal = ½KA2 = 25 J.
The equation for the acceleration of an object undergoing SHM is has the form a(t) = -amaxcos(ωt+Δ), where amax = ω2A and ω = 2π/T. Examining the graph, we see that the period is T = 0.030 s, so ω = 200π/3 s-1. Also the maximum acceleration is 0.040 m/s2. From this we deduce that A = amax/ω2 = (0.040 m/s2)/(200p/3 s-1)2 = 9.119 × 10-6 m. Furthermore, the maximum velocity is vmax = ωA = amax/ω = 1.9099 × 10-4 m/s.
The position equation for SHM is x(t) = Acos(ωt + Δ). We just found ω and A so all that is left to do is to find Δ. Again we look at the value of the graph at t = 0 which is a(0) = 0.035 m/s. So we have 0.035 = -0.040cos(Δ), or cos(Δ) = -0.875. There are two angles on the unit circle that satisfy this, Δ1 = 151.04° and Δ2 = -151.04°. We examine the unit circle for these possibilities
Now the only difference in the x components of these two graphs, is that Δ1 gives a negative velocity while Δ2 gives a positive velocity. To determine the sign of the velocity at t = 0 from the acceleration graph above, we need to integrate the area under the curve near t = 0. That area is positive, so the velocity needs to be positive. Thus the correct phase angle is Δ2 = -151.04° = -2.64 rad. Our equation is x(t) = (9.119 × 10-6 m)cos(200πt/3 - 2.64).
The spring constant is given by ω2 = K/m, so K = (200π/3 s-1)2(2.00 kg) = 87730 N/m.
The total energy is Etotal = ½KA2 = 3.65 × 10-6 J.
An examination of our relationships for SHM indicates
that the velocity of a particle is at a maximum as it goes through
equilibrium. So the problem has given use the maximum velocity.
As well we have vmax = ωA,
and ω = [K/m]½, so
i. The fact that we are looking for w indicates an SHM problem. It is not the simple one spring and block system we studied since there are two springs. However energy methods let us handle everything since we know Ef = Ei, hence we have
Solving for v yields 
.
ii. Now this is the maximum velocity of the block and we know for SHM that
vmax = ωA. So 
.
iii. For a single spring we know 
. Since
ω and M are the same, Keffective =
2K. We replace the two springs with a single spring with twice the spring constant.
i. The fact that we are looking for ω indicates an SHM problem. It is not that simple a spring and block system since there is also rolling. However energy methods let us handle everything since we know Ef = Ei, hence we have
For a disk, I = ½Mr2. Since it is rolling, ωrolling = v/r. The above equation reduces to
Solving for v yields 
.
ii. Now this is the maximum velocity of the block and we know for SHM that
vmax = ωA. So
.
The problem tells us that the block loses contact with the spring. That means the normal acting on the block goes to zero. Typically, for problems involving forces we draw the FBD and apply Newton's Second Law. However, we have always avoided doing this with springs because we are dealing with a non-constant force and a non-constant acceleration. We don't have this difficulty in this problem because we are told that the spring is at maximum extension, and thus we know that the acceleration is amax = ω2A downwards.
 |
|
| ΣFy = may | |
| N - mg = -mω2A |
The equation we have is
Setting N = 0, and rearranging yields
Our relationships for SHM tell us that ω
= [K/m]½, so
Examining the motion we see that the object swings for half the
motion on a string of length L, and the other half as a string
of length L/n. From the period of a pendulum, we know how long
each of these motion are. Therefore the total time is
Questions? mike.coombes@kwantlen.ca
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