Chapter 15 Oscillatory Motion I
Level : AP Physics
Instructor : Kim
Introduction
A very special kind of motion occurs when the force acting on a body is proportional to the displacement
of the body from some equilibrium position. If this force is always directed toward the equilibrium
position, repetitive back-and-force motion occurs about this position. Such motion is called periodic
motion, or harmonic motion, oscillation or vibration.
Examples of periodic motion; a block attached to a spring, the swinging of a child on a playground, the
motion of a pendulum, and the vibrations of a stringed musical instrument, etc . . .
The ideal motion is called simple harmonic motion, where there is no loss in energy
Ideal Spring
When the block is displaced a small distance x from equilibrium, the spring exerts on the block a force
that is proportional to the displacement and given by Hook’s law
Fs = –kx : Hook’s Law
We call this a restoring force because it is always directed toward the equilibrium position therefore
opposite the displacement. That is, if a force is applied to a spring, the spring exerts an restoring force that
is always opposite in direction
Fs
∑F = Fs + Fapp = 0
=> Fapp = -Fs
=> Fapp = -(-kx)
Fapp =kx
Fapp
Fapp=kx
Q1) An object of mass 9kg is hung on a spring, causing the spring to
stretch by x=0.0191m. Then determine the force needed to stretch the
spring by 0.0508m.
a) 194N
b) 235N
c) 276N
d) 310N
x
Fg =mg
(b)
(a)
F(N)
Q2) The graph shows the force F that an archer applies to the spring
160
of a long bow versus the string’s displacement x. Drawing back this
bow is analogous to stretching a spring. From the data in the graph
determine the effective spring constant of the bow
a) 324.5N
b) 397.1N
c) 533.6N
d) 666.7N
x(m)
0.24
Simple Harmonic Motion and Position(or displacement) Function over time
When a block of mass m is attached to a spring that is
neither stretched nor compressed, the block is at the position
x=0, called the equilibrium position of the system. If the
block is disturbed from its equilibrium position, the block
will oscillate back and forth.
FP
x=0
For example, assuming the surface is frictionless, if the block is given a 'instantaneous' push(FP) to the
right, then the block will oscillate.
If we record the position of the block through time and plot in on a x-t graph, we can see the following;
v=0
Fs
x(m)
1
t(s)
1.57
x=0
x=1m
-1
Assume the block momentarily stops at position x=1m and the time it takes to reach that position is t=1.57s
vmax
Fs= 0
x(m)
1
t(s)
1.57
x=0
3.14
-1
The block returns to the origin, reaching maximum speed the moment it passes the origin
x(m)
1
v=0
Fs
4.71
t(s)
1.57
x= –1m
x=0
3.14
-1
If the surface is frictionless, then the block will be compressed until it reaches x= –1m. (maximum compression)
vmax
Fs= 0
x(m)
1
4.71
6.28
t(s)
1.57
x=0
-1
3.14
Q4) A block is attached to spring and set in simple harmonic motion. Surface is
frictionless. Which of the following function best describes the position(x)
function of the block as a function of time?
x
a) x(t) = 2t
b) x(t) = t2
(c) x(t) = 3√𝑡
d) x(t) = sint
x=0
xmax
t
Hence, the displacement function that best describes the motion over time can be expressed as
x(t) = sint
where the maximum stretch or compression is x= ±1m. If the block is stretched or compressed more, like
x=2m, then x(t) = 2sint
However, since our observation begins according to our choice and if we
happened to start observing when the block is at the maximum stretch from the
equilibrium position, then the displacement function can be expressed as
𝜋
x(t) = 2sin(t + 2 ), where π/2=1.57
x=0
xmax
x
or
x(t) = 2cost
t
Hence, choosing cosine or sine function depends on when we start our observation
Q5) Which of the following function is equal to x(t) = cos(t - π)? There are two answers
𝜋
𝜋
a) x(t) = sint
b) x(t) = sin(t - 2 )
c) x(t) = cos(t - 2 )
d) x(t) = – cost e) x(t) = cost
Q6) Which graph represents the position x(t) function? Connect by drawing a line
x
x(t) = cos(t+3π/2)
t
x
x(t) = cos(t+π/2)
t
x
x(t) = cos(t - π)
t
Newton's 2nd law and Position Function for Simple Harmonic Motion
Applying Newton’s 2nd law to the motion of the block, we obtain
Fs = –kx= ma
𝑘
a = −𝑚𝑥
𝑑2 𝑥
𝑑𝑡 2
𝑘
= −𝑚𝑥
solving the above equation gives
x(t)=Acos(wt+φ)
where w2=k/m, k is the spring constant, m is mass of the block
x(t)=Acos(wt+φ) is the displacement function of a particle exhibiting simple harmonic motion. That is,
the function provides the position of the particle from the equilibrium at any time
- ‘A’ is the amplitude, which represents the maximum displacement of the particle either the positive or
negative x direction.
- ‘w’ is the angular frequency of the motion and has units of rads /s
- The constant angle φ is the phase constant. It is determined by initial displacement and velocity of the
particle
The quantity (wt + π) is called the phase of the motion and is useful in comparing the motion of two
oscillators.
Note that the trigonometric function Acos(wt+φ) is periodic and repeats itself every time ‘wt’ increases by
2πrads. Here we can define the period T, since period T of the motion is the time it takes for the
particle to go through one full cycle or one oscillation.
Hence, the value of x at time t equals the value of x at time t+T.
wt + φ + 2π = w(t +T) + φ
𝟐𝝅
Hence T = 𝒘
In rotational motion, recall that the angular speed w was defined as w=Δθ/Δt. If an object completes one
cycle, then Δθ=2π and Δt=T. So w=Δθ/Δt=2π/T, then angular speed, which is also often called angular
frequency can be written as
2𝜋
𝑤=
𝑇
The inverse of the period is called the frequency f of the motion. The frequency represents the number of
oscillations that the particle makes per unit time:
1
𝑤
𝑓 = 𝑇 = 2𝜋 and 𝑤 = 2𝜋𝑓
The unit of f are cycles per second, that is s-1 or hertz(Hz).
𝒌
𝒎
Since w=√𝒎 , the period can also be expressed as T=2π√ 𝒌 and f =
Summary
Fs=–kx
and
𝟏
𝒌
√
𝟐𝝅 𝒎
Fapp=kx
The position x of a simple harmonic motion: x(t)=Acos(wt+φ)
𝑘
𝑚
A is the amplitude, w is the angular frequency where w=√ , φ is the phase constant
T is the period where 𝑇 =
2𝜋
𝑤
𝑚
𝑘
=2𝜋√
1
𝑤
1
𝑘
f is the frequency where 𝑓 = 𝑇 = 2𝜋 = 2𝜋 √𝑚
Q7) An object undergoes simple harmonic motion of amplitude A. Through what total distance does the
object move during one complete cycle of its motion?
(a) A/2
(b) A
(c) 2A
(d) 4A
Q8) If an object takes 2seconds to complete one cycle, what are the period, frequency and angular
frequency? Check your answer below
Ans) 2s, 0.5Hz, 3.14rad/s
Velocity and Acceleration Function for Simple Harmonic Motion
We can obtain the linear velocity of a particle undergoing simple harmonic motion by differentiating
equation x(t)=Acos(wt+φ)
The speed of the particle is v(t)=
𝒅𝒙
𝒅𝒕
= -wAsin(wt + φ ), where vmax=wA (compare with v=rw)
The acceleration of the particle is a(t) =
ar=v2/r=w2r)
𝒅𝒗
=
𝒅𝒕
-w2Acos(wt + φ ), where amax=w2A(compare with
Displacement vs Velocity vs Acceleration
A block attached to a spring on a frictionless surface is pulled a certain distance away from the
equilibrium and then released. The block has maximum displacement the instant it was released. At that
moment, the speed is minimum and acceleration is maximum.
(a) Displacement vs time graph is shown. x(t) is maximum the
moment the block is being released. x(t) is minimum the
moment the block passes the equilibrium position
x(t)
A
t
x(t)=Acos(wt)
(b) Velocity vs time graph is shown. v(t)=0 the moment the block
is being released. v(t) is maximum in the negative direction the
moment the block passes the equilibrium position
𝑑𝑥
v(t)= 𝑑𝑡 = -Awsin(wt)
-A
v(t)
wA
t
-wA
w2A
a(t)
(c) Acceleration vs time graph is shown. a(t) is maximum in
negative direction the moment the block is being released.
a(t)=0 the moment the block passes the equilibrium position
𝑑𝑣
a(t)= 𝑑𝑡 = -Aw2cos(wt)
-w2A
Calculators must be in radians!!!!!!
Or plunge into forever darkness~
Q9) The displacement of a particle at t=0.25s is given by the expression x(t)=(4.00m)cos(3πt + π), where
x is in meters and t is in seconds. Determine
meters
(a) the frequency and period of the motion
Ans) 1.5Hz, 0.67s
(b) the amplitude of the motion
Ans) 4m
(c) the phase constant
Ans) π
(d) the displacement of the particle at t=0.25s.
Ans) 2.83m
(e) Plot a displacement vs time graph of the motion below. Plot your graph manually and then check
calculator
x(m)
t(s)
Q10) An object oscillates with simple harmonic motion along the x axis. Its displacement from the origin
varies with time according to the equation
𝜋
x = (4.00m) cos( πt + )
4
where t is in seconds and the angles in the parentheses are in radians.
(a) Determine the amplitude, frequency and period of the motion
Ans) 4m, 0.5Hz, 2s
(b) Calculate the speed and acceleration of the object at time t=1s.
Ans) 8.89m/s (right) 27.9m/s2
(c) Determine the maximum speed and at what time does the object have maximum speed?
Ans) 12.6m/s, vmax at t = 0.25s, 1.25s, 2.25s , . . . .
(d) Plot a displacement vs time graph of the motion below.
x(m)
t(s)
(e) Find the displacement of the object between t = 0 and t =1s and the distance traveled during those
time.
Ans) ∆x= ‒5.66m, d=8m
Q11) A 0.8kg object is attached to one end of a
spring and the system is set into simple harmonic
motion. The displacement x of the object as a
function of time is shown. With the aid of this data,
determine the
(a) amplitude A of the motion
Ans) 0.08m
x(m)
0.08
1.0
2.0
3.0
4.0
time(s)
0.08
(b) the angular frequency w and frequency f
Ans) 1.57rad/s, 0.25Hz
(c) the spring constant k
Ans)1.97N/m
(d) the maximum displacement from the equilibrium position xmax
(e) At what time(s) is the object at the equilibrium position?
Ans) 0.08m
Ans) t=0,2,4, . .
(f) Express the displacement of the motion as a function of time x(t)
𝜋
𝜋
𝜋
2
2
2
Ans) x(t)=0.08cos( 𝑡 − ) or 0.08sin( 𝑡)
(g) Find the position of the object at t=0.2, 0.5, 2.4, 3.0 seconds
Ans) 0.0247m, 0.0565m, ‒0.047m, ‒0.08m
(h) What is the maximum speed of the object? Also find the expression for speed as a function of time v(t)
𝜋
𝜋
2
2
Ans) 0.1256m/s, v(t)= ‒0.04πsin( 𝑡 − )
(i) Find the speed of the object at t=0.4 , 1.5 and 3.2 seconds.
Ans) 0.102m/s, ‒0.089m/s, 0.039m/s
(j) Find the an expression for acceleration as a function of time a(t) and find the acceleration at t=0.4 ,
1.5, 3.2, 3.9, 3.99, 3.999 seconds
𝜋
𝜋
Ans) a(t)= ‒0.02π2cos( 2 𝑡 − 2 ), ‒0.116m/s2, ‒0.14m/s2, 0.188m/s2, 0.0316m/s2, ‒0.0031m/s2, approach ‘0’