Download Handout 5: Oscillatory motion Simple harmonic motion Simple

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Handout 5: Oscillatory motion
Simple harmonic motion
Simple harmonic motion (SHM) is periodic
motion with constant period or frequency. It is the
motion where acceleration is proportional to
displacement in the opposite direction. The
acceleration 𝑎 and displacement 𝑥 are related by
𝑎 = −𝜔2 𝑥,
Figure 1: Mass-spring system
where 𝜔 is the angular frequency of the motion. The
period and frequency can be found from 𝜔.
In Figure 1, a mass 𝑚 on a smooth floor is
attached to one end of a spring whose other end is fixed.
The spring has spring constant 𝑘. The mass is displaced
to the right and let go; it performs an oscillatory motion
along x-axis. When the displacement is 𝑥, there is a force
– 𝑘𝑥 acting on the mass according to Hooke’s law. This
force in the spring is a restoring force. The equation of
motion is
𝑚𝑥 = −𝑘𝑥
→
𝑥 = −
It can be seen that 𝜔2 = 𝑘/𝑚 or 𝜔 =
the period of oscillations
𝑇 =
𝑘
𝑥.
𝑚
𝑘/𝑚. Therefore,
2𝜋
𝑚
= 2𝜋
.
𝜔
𝑘
Example A mass 𝑚 is suspended from a vertical spring
whose other end is fixed. The spring constant is 𝑘. Show
that the period of motion is
𝑇 = 2𝜋
𝑚
.
𝑘
Example A mass 𝑚 is attached by two springs with
different spring constants 𝑘1 and 𝑘2 . Show that the mass
undergoes SHM and find the formula for the period of
the motion.
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Example A wooden block of mass 𝑚 and cross-section
area 𝐴 is partially immersed in a liquid with density 𝜌.
The block is displaced downwards by a distance and
released. Show that the block performs SHM and
determine the angular frequency of the motion.
Solution to SHM equation
Let’s go back to the mass-spring system as shown
in Figure 1. The mass is displaced by distance 𝐴 and let
go. Therefore, 𝐴 is the maximum displacement which is
known as the amplitude of the oscillations. The
equation of motion is
𝑑2 𝑥
= −𝜔2 𝑥,
𝑑𝑡 2
𝜔 =
𝑘
.
𝑚
We wish to find displacement 𝑥(𝑡) and velocity 𝑣(𝑡) as
functions of time 𝑡. The general solution is
𝑥 𝑡 = 𝐶1 sin 𝜔𝑡 + 𝐶2 cos 𝜔𝑡,
where 𝐶1 and 𝐶2 are constants. The above equation can
be expressed in the form
𝑥 𝑡 = 𝐴 sin(𝜔𝑡 + 𝛿),
where 𝐴 is the amplitude and 𝛿 is a phase angle. The
shape function 𝑥 𝑡 is illustrated in Figure 2. There are
two cases to consider

At 𝑡 = 0, 𝑥 = 0, we find that 𝛿 = 0:
𝑥 𝑡 = 𝐴 sin 𝜔𝑡.
 At 𝑡 = 0, 𝑥 = 𝐴, we find that 𝛿 = 𝜋/2:
𝑥 𝑡 = 𝐴 cos 𝜔𝑡.
From displacement 𝑥 𝑡 = 𝐴 sin(𝜔𝑡 + 𝛿), one
can differentiate to obtain velocity 𝑣(𝑡) and acceleration
𝑎 𝑡 :
Figure 2: Displacement against time
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𝑣 𝑡 = 𝜔𝐴 cos(𝜔𝑡 + 𝛿) ,
𝑎 𝑡 = −𝜔2 𝐴 sin(𝜔𝑡 + 𝛿)
The graphs of 𝑣(𝑡) and 𝑎 𝑡 are shown in Figure 3. It
should be noted that the maximum velocity 𝑣𝑚𝑎𝑥 and
maximum acceleration 𝑎𝑚𝑎𝑥 are given by
𝑣𝑚𝑎𝑥 = 𝜔𝐴,
𝑎𝑚𝑎𝑥 = 𝜔2 𝐴.
Example A particle performs simple harmonic motion
where acceleration 𝑎 in ms-2 is a function of
displacement 𝑥 in m according to 𝑎 = −2.65𝑥. The
amplitude of the motion is 0.35 m and the particle is at
𝑥 = 0 when time 𝑡 = 0 s. Determine
a) Angular frequency
b) Maximum speed
c) Displacement when 𝑡 = 2.5 s
Example For a SHM motion with solution
𝑥 𝑡 = 𝐴 sin(𝜔𝑡 + 𝛿),
show that velocity 𝑣(𝑥) as a function of 𝑥 is given by
𝑣 = 𝜔 𝐴2 − 𝑥 2 .
Example A particle 𝑃 moves along the 𝑥-axis. At time 𝑡
seconds, its displacement, 𝑥 meters, from the origin 𝑂 is
given by 𝑥 = 5 sin
𝜋
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𝑡 . Points 𝐴 and 𝐵 are on 𝑥-axis
such that 𝑂𝐴 = 2 m and 𝑂𝐵 = 3 m. Find the time taken
by 𝑃 to travel directly from 𝐴 to 𝐵.
Figure 3: Displacement, velocity and
acceleration against time
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Example A particle P moves on the x-axis with simple
harmonic motion about the origin O as centre. When P is
a distance 0.04 m from O, its speed is 0.2 ms–1 and the
magnitude of its acceleration is 1 ms–2. The amplitude of
the motion is A meters.
a) Find the period of oscillations.
b) Find the amplitude A.
c) Find the total time, within one complete
oscillation, that distance OP is greater than A/2.
Damped harmonic motion
For simple harmonic motion, the total energy of
the system is constant; the amplitude of the oscillations
is also constant. If damping force is taken into account,
the energy is lost and therefore the amplitude decreases
with time. Such motion is called damped harmonic
motion. The damped harmonic motion can be modeled
as shown in Figure 4. Apart from the usual SHM setup
which consists of mass and spring, the mass is
connected to a light vane immersed in a liquid. As the
mass moves at speed 𝑥, there is a drag force – 𝑏𝑥 acting
on the vane. The constant 𝑏 is related to the viscosity of
the liquid.
The equation of motion for the mass becomes
𝑚𝑥 = −𝑏𝑥 − 𝑘𝑥.
Figure 4: Damped harmonic motion
setup
For b not too large, the solution to the above equation is
given by
𝑥 𝑡 = 𝐴0 𝑒 −(𝑏
2𝑚 )𝑡
sin 𝜔𝑡 + 𝛿 ,
𝜔 =
𝑘
𝑏2
−
,
𝑚 4𝑚2
where 𝐴0 is a constant. The value of b has to be small
enough to keep 𝜔 real. The term in front of sine function
represents the amplitude A of the motion which is
decreasing with time (Figure 5):
𝐴 = 𝐴0 𝑒 −(𝑏
2𝑚 )𝑡
.
Hence, the constant 𝐴0 is the initial amplitude. This type
of damping is called underdamping. If the value of 𝑏 is
Figure 5: Decay in amplitude of
underdamped motion.
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too large, the damping may become critical damping or
overdamping. The comparison is illustrated in Figure 6.
For critically damped motion, the value of b is high,
there is no oscillation and 𝑥 reaches zero in short time.
For overdamped motion, the value of b is even higher;
there is also no oscillation and 𝑥 reaches zero in a long
time.
Example A particle of mass 𝑚 = 0.1 kg is attached to a
spring with constant 𝑘 = 250 Nm-1 and undergoes a
damped harmonic motion with drag force with
coefficient 𝑏 = 0.1 kg s-1.
a) Evaluate the period of the oscillations.
b) Find time at which the amplitude is halved.
c) By the time in b), how many cycles have passed?
Figure 6: Types of damped
harmonic motion
Example For a damped harmonic motion of an object
with drag force 𝑓 = −𝑏𝑣 where 𝑏 is a constant and 𝑣 is
the speed of the object, show that the rate of change in
energy of the system is equal to −𝑏𝑣 2 .
Forced oscillations and resonance
Mass-spring system has a natural frequency
𝑓0 =
1 𝑘
,
2𝜋 𝑚
(assume no damping). That means whenever the mass
oscillates, it will do so with the natural frequency. In
Figure 7, the mass is forced to oscillate by an oscillatory
driving force by hand on the top end of the spring. The
driving force has a driving frequency 𝑓. If the driving
frequency is equal to the natural frequency, the result is
Figure 7: Driven oscillations. The
mass is forced to oscillate.
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that the mass will oscillate with large amplitude. This
phenomenon is known as resonance.
At resonance, the amplitude of the oscillation is
large. The size of the amplitude at resonance depends on
the degree of damping as shown in Figure 8. It is
important to look at the phase difference 𝛿 between
driving force (hand) and the oscillating mass.

When 𝑓 ≪ 𝑓0 , we find 𝛿 = 0
 hand and mass are in phase.
 When 𝑓 = 𝑓0 , we find 𝛿 = 𝜋/2
 hand and mass out of phase by 𝜋/2.
 When 𝑓 ≫ 𝑓0 , we find 𝛿 = 𝜋
 hand and mass out of phase by 𝜋.
Resonance occurs in other systems too. A system
has its own natural frequencies. When the frequency of
the driving force matches one of the natural frequencies,
resonance occurs and it can be destructive. Examples
are
 Glass-breaking sound The frequency of sound
matches the natural frequency of the glass, the
result is strong vibrations of the glass until it
breaks.
 Loud speakers Boom or buzz is heard when a
musical note happens to coincide with the
resonant frequency of the speaker cone.
 Tuned radio A tuned circuit in radio receiver
responds strongly to a particular frequency.
 Airplane’s wings If frequency of vibrations of the
engines of particular airplane is right to resonate
with the natural frequencies of the wings. Large
oscillations build up and the wings may fall off.
 Tacoma Narrows Bridge (Figure 9) The wind
across the bridge was turbulent and vortices
were formed with regular frequency. This
frequency matched the natural frequency of the
bridge. The result was violent vibrations of the
bridge which collapsed eventually.
Figure 8: Graphs of amplitude against
driving frequency.
Figure 9: Tacoma Narrows Bridge