Download Connection between Simple Harmonic Motion and Uniform Circular Motion in Physics and more Study notes Physics in PDF only on Docsity!
Section 35 – SHM and Circular Motion
“What do objects do?” and “Why do they do it?” Objects sometimes oscillate in simple
harmonic motion. In the last section we looked at mass vibrating at the end of a spring. We applied the
Second Law to derive the equations of motion for SHM. In the process, we seemed to be using the idea
of angular frequency just as we did when we looked at uniform circular motion. In addition, the
equations of motion for SHM look every similar to the equations of motion for uniform circular motion.
In this section, we will investigate the connection between SHM and uniform circular motion.
Next, we’ll continue to build our understanding of SHM by looking at the oscillatory motion of a
simple pendulum. We’ll discover that it also is SHM under certain conditions.
Section Outline
The Connection Between Uniform Circular Motion and SHM
Energy in SHM
The Simple Pendulum 1. The Connection Between Uniform Circular Motion and SHM
We have been using the idea of angular frequency as we did when
we discussed circular motion. There must be some connection, so
let’s investigate. At the right is an object going in a circle on a
rotating turntable. Just behind the object is a screen where the
shadow of the object can be seen. The shadow moves back and
forth as the object goes in a circle. The shadow appears to be in
SHM.
At the right is a sketch of the object in uniform circular motion. It has a
centripetal acceleration, a tangential velocity, and a position vector all
shown. If we redraw the three vectors with their tails at the origin we can
imagine all three spinning as the object rotates. Finding the x-components
of each – the position, velocity, and acceleration of the shadow,
x = r cos θ , v
x
= − v sin θ , and a x
= − a cos θ.
The tangential velocity is related to this angular velocity,
v = ωr.
Also, the centripetal acceleration is related to the angular velocity,
a =
v
2
r
(ω r)
2
r
⇒ a = ω
2 r.
Substituting, we get,
x = r cos θ , v
x
= − ω r sin θ , and a x
2
r cos θ.
Notice, just like SHM we have a x
2 x. The angle θ changes with
time. We can write this using the definition of angular frequency,
ω ≡
dθ
dt
⇒ dθ
= ωdt
⇒ θ = ωt + δ.
Now we see another way of looking at the phase angle, δ, as just an integration constant.
Finally, we can write the x-components of the, position, velocity and acceleration for the oscillating
shadow as a function of time,
x
y
a
r
v
θ
x
y
a
r v
θ
θ
θ
x ( t ) = r cos( ω t + δ) , v ( t ) = − ω r sin( ω t + δ) , and a ( t ) = − ω
2 r cos( ω t + δ).
These are the same as the SHM equations of motion with A instead of r. The x component of the motion
of an object in uniform circular motion is SHM. That explains why we keep talking about angular
frequencies!
Example 35.1: A 500g mass rests in equilibrium at the end of a horizontal spring with spring
constant 9.80N/m. The mass is given a sharp kick resulting in an initial velocity of 0.443m/s to
the right. (a)Sketch the initial position, velocity, and acceleration vectors as if the object were in
circular motion. Find (b)the location of the equivalent object
in circular motion, (c)the phase angle, and (d)the equation for
v(t).
Given: k = 9.80N/m, m = 0.500kg, v(0) = 0.443m/s, and
x(0) = 0.
Find:
r =? ,
v =? ,
a =? , d = ?, and x(t)=?
(a)We are given that the velocity vector is to the right and the
initial position is zero. Therefore, the x-component of the
velocity must be at a maximum and point to the right. The x-
component of the position and acceleration vectors must be
zero. The acceleration must point toward the center of the
circle and the position must point outward. The answer then
is in the sketch at the upper right.
(b)The equivalent object in circular motion with the vectors
pointing the right direction must be as shown at the right.
(c)Looking at the circle, the phase angle must be 270˚ or
3 π
2
(d)Using the appropriate equation of motion for circular
motion, the equation for the position as a function of time is
v ( t ) = − v o
sin( ω t + δ ).
The angular frequency for a spring is,
ω =
k
m
rad
s
So,
v ( t ) = −0.443sin(4.43 t +
3 π
2
Note this results in v(0) = +0.443m/s as required.
x
y
a
r
v
θ
x
y
θ
a r
v
A “simple” pendulum consists of a very light string with a concentrated mass at the
end. The forces on the mass at the end are gravity and the tension. However, the
tension exerts no torque about the top of the string. Applying the Second Law for
Rotation,
Στ p
= Iα ⇒ −mg sinθ = m
2 α ⇒ α = −
g
sinθ.
Since we are looking at rotational motion, we are checking to see if the angular
acceleration is equal to the negative of some constants multiplied by the angular
position. Sadly, this is not the case for the simple pendulum because we have a sinθ
instead of just θ. However, for small angles,
sinθ ≈ θ ⇒ α ≈ −
g
θ.
This is the SHM equation, the acceleration (angular, in this case) is equal to minus
some constants times the position (again, angular). This is the same equation we got for the motion of
the mass on the end of spring, except that θ replaces x. In other words, the equations of motion for the
angle, θ, will be the simple harmonic motion equations with an angular frequency equal to the root of
the constants so long as the angle is small.
Angular Frequency of a Simple Pendulum ω =
g
Example 35.2: The pendulum in a grandfather clock must have a period of 2.00s so that each
swing moves the second hand twice. Find the length of the pendulum.
Given: T = 2.00s
Find: =?
The angular frequency of a simple pendulum is,
ω =
g
It is related to the period,
ω = 2 πf =
2 π
T
⇒ T =
2 π
ω
⇒ T = 2 π
g
gT
2
4 π
2
Plugging in the numbers,
2
4 π
2
= 0.993 m (^).
This explains why all pendulums in grandfather clocks are about this size.
Fg
P
θ
Ft
Section Summary
Why do objects do what they do? We have been building our understanding of simple harmonic
motion. We have learned that an object with an acceleration that is equal to minus the product of some
constant and the position is in SHM an obeys the SHM equations of motion,
a(x) = −ω
2 x
v(x) = ±ω A
2 − x
2
x(t) = A cos( ω t + δ)
v(t) = −ωA sin (ω t + δ)
a(t) = −ω
2 A cos (ω t + δ)
where A is the amplitude of the motion, ω is the angular frequency, and δ is the phase angle. The
angular frequency will be equal to the root of the constants.
We examined the connection between circular motion and SHM. SHM is the motion of the shadow of
an object in uniform circular motion. In other words, the equations of motion for the x-component of
uniform circular motion are identical to the equations of motion for SHM.
With the knowledge above, we look at the oscillations of a simple pendulum and found that they are
indeed SHM with an angular frequency given by,
ω =
g
