Oscillations Unit 7. Lesson 1 : Simple Harmonic Motion F s = -kx (Hooke’s Law) F s is a restoring...
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Transcript of Oscillations Unit 7. Lesson 1 : Simple Harmonic Motion F s = -kx (Hooke’s Law) F s is a restoring...
![Page 1: Oscillations Unit 7. Lesson 1 : Simple Harmonic Motion F s = -kx (Hooke’s Law) F s is a restoring force because it always points toward the equilibrium.](https://reader030.fdocuments.in/reader030/viewer/2022013115/56649e755503460f94b7611f/html5/thumbnails/1.jpg)
Oscillations
Unit 7
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Lesson 1 : Simple Harmonic Motion
Fs = -kx
(Hooke’s Law)
Fs is a restoring force because it always points toward the
equilibrium position (x = 0)
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Applying Newton’s Second Law :
F = max
-kx = max
ax = -k
mx
Simple Harmonic Motion
An object moves with simple harmonic motion whenever its acceleration is proportional to its
position and is oppositely directed to the displacement from equilibrium.
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Example 1
A block on the end of a spring is pulled to position x = A and released. In one full cycle of its motion, through what
total distance does it travel ?
_____ A/2
_____ A
_____ 2A
_____ 4A
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a =dv
dt=
d2x
dt2Since ,
d2x
dt2= -
k
mx
Notice that the acceleration of the particle in SHM is not constant. It
varies with position x.
If we call k/m = 2,
d2x
dt2= -2x
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We need a function x(t) whose second derivative is the same as the original function
with a negative sign and multiplied by 2.
d2x
dt2= -2x
(second-order differential equation)
x(t) = A cos(t + )
One possible solution is :
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Proof
dx
dt= A cos(t + )
d
dt= -A sin(t + )
d2x
dt2= -A sin(t + )
d
dt= -2A cos(t + )
d2x
dt2= -2x
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Amplitude (A) : maximum value of the position of the particle in either the positive or negative direction.
x(t) = A cos(t + )
Angular Frequency () : number of oscillations per second.
Since k/m = 2,
= km
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= 0
Phase Constant () : initial phase angle. This is determined by the position of the particle at t = 0.
If particle is at maximum position x = A at t = 0, the
phase constant = 0.
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Pen traces out cosine curve x(t) = A cos(t + )
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Period (T) : the time interval required for the particle to go through one full cycle of its motion.
T =2
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Frequency (f) : the inverse of period. The number of oscillations that the
particle undergoes per second.
f =1
T
Since T = 2/,
f =2
The Hertz (Hz) is the SI unit for frequency.
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Since f = /2,
= 2f =2T
Since = k/m and T = 2/ ,
T = 2 m
k
f =k
m
1
2
depend only on mass and
spring constant
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Velocity in Simple Harmonic Motion
dx
dt= A cos(t + )
d
dt= -A sin(t + )
v = -A sin(t + )
Acceleration in Simple Harmonic Motion
d2x
dt2= -A sin(t + )
d
dt= -2A cos(t + )
a = -2A cos(t + )
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Maximum Speed in Simple Harmonic Motion
vmax = +/- A
Since sine and cosine oscillate between +/- 1,
vmax =km
A (magnitude only)
Maximum Acc. in Simple Harmonic Motion
amax = +/- 2A
amax =km A (magnitude only)
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position vs. time
velocity vs. time
acceleration vs. time
phase difference is /2 rad or 90o
when x at max or min, v = 0
when x = 0, v is max
phase difference is rad or 180o
when x at max, a is max in opposite direction
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Example 2
An object oscillates with simple harmonic motion along the x-axis. Its position varies
with time according to the equation
x = (4.00 m) cos(t + /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.
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b) Calculate the velocity and acceleration of the object at any time t.
c) Using the results of part b, determine the position, velocity, and acceleration of the object at t = 1.00 s.
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d) Determine the maximum speed and maximum acceleration of the object.
e) Find the displacement of the object between t = 0 and t = 1.00 s.
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Example 3A 200 g block connected to a light spring for
which the force constant is 5.00 N/m is free to oscillate on a horizontal, frictionless surface. The block is displaced 5.00 cm from equilibrium and
released from rest, as shown below.
a) Find the period of its motion.
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b) Determine the maximum speed of the block.
c) What is the maximum acceleration of the block ?
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d) Express the position, speed, and acceleration as functions of time.
e) The block is released from the same initial position, xi = 5.00 cm, but with an initial velocity of vi = -0.100 m/s. Which parts of the solution change and what are the new answers for those that do change ?
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A 2 kg block is dropped from a height of 0.45 m above an uncompressed spring, as shown above. The spring has an elastic constant of 200 N/m and negligible mass. The block strikes the end of the
spring and sticks to it.
Example 4 : AP 1989 # 3
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a) Determine the speed of the block at the instant it hits the end of the spring.
b) Determine the period of the simple harmonic motion that ensues.
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c) Determine the distance that the spring is compressed at the instant the speed of the block is maximum.
d) Determine the maximum compression of the spring.
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e) Determine the amplitude of the simple harmonic motion.
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Example 5 : AP 2003 # 2
An ideal spring is hung from the ceiling and a pan of mass M is suspended from the end of the
spring, stretching it a distance D as shown above. A piece of clay, also of mass M, is then
dropped from a height H onto the pan and sticks to it. Express all algebraic answers in terms of
the given quantities and fundamental constants.
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a) Determine the speed of the clay at the instant it hits the pan.
b) Determine the speed of the pan just after the clay strikes it.
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c) Determine the period of the simple harmonic motion that ensues.
d) Determine the distance the spring is stretched (from its initial unstretched length) at the
moment the speed of the pan is a maximum. Justify your answer.
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The clay is now removed from the pan and the pan is returned to equilibrium at the end of the spring. A rubber ball, also of mass M, is dropped from the same height H onto the pan, and after the collision
is caught in midair before hitting anything else.
e) Indicate below whether the period of the resulting simple harmonic motion of the pan is greater than, less than, or the same as it was in part c.
____ Greater than ____ Less than ____ The same as
Justify your answer.
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Lesson 2 : Energy in Simple Harmonic Motion
KE of Block
KE = ½ mv2
Since v = -A sin(t + ),
KE = ½ m2A2 sin2(t + )
Elastic PE Energy Stored in Spring
U = ½ kx2
Since x = A cos(t + ),
U = ½ kA2 cos2(t + )
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Total Mechanical Energy of Simple Harmonic Oscillator
E = KE + U
E = ½ m2A2 sin2(t + ) + ½ kA2 cos2(t + )
E = ½ kA2 [sin2(t + ) + cos2(t + )]
Since 2 = k/m,
Since sin2 + cos2 = 1,
E = ½ kA2
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E = ½ kA2
The total mechanical energy of a simple harmonic oscillator is a constant of the
motion and is proportional to the square of the amplitude.
U is small when KE is large, and vice versa.
KE + U = constant
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E = ½ kA2
Since E = KE + U,
½ kA2 = ½ mv2 + ½ kx2
Solving for v,
v = +/-km
(A2 – x2)
Since =
km
,
v = +/- A2 – x2
velocity of simple
harmonic oscillator
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Example 1
The amplitude of a system moving in simple harmonic motion is doubled.
Determine the change in the
a) total energy
b) maximum speed
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c) maximum acceleration
d) period
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Example 2
A 0.500 kg cart connected to a light spring for which the force constant is
20.0 N/m oscillates on a horizontal, frictionless air track.
a) Calculate the total energy of the system and the maximum speed of the cart if
the amplitude of the motion is 3.00 cm.
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b) What is the velocity of the cart when the position is 2.00 cm ?
c) Compare the kinetic and potential energies of the system when the position is 2.00 cm ?
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Lesson 3 : Comparing Simple Harmonic Motion with Uniform Circular Motion
As the turntable rotates with constant angular speed, the shadow of the ball moves back and forth in simple harmonic motion.
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Reference Circle
Simple harmonic motion along a straight line can be represented by the projection of uniform circular motion along a diameter of a reference circle.
t = 0
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t > 0
Q is the projection
of P
x coordinate is x(t) = A cos(t + )
Since = t =
Point Q moves with simple
harmonic motion
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What is the amplitude and phase constant (relative to an x axis to the right) of the simple harmonic motion of
the ball’s shadow ?
_____ 0.50 m and 0
_____ 1.00 m and 0
_____ 0.50 m and
_____ 1.00 m and
Example 1
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While riding behind a car traveling at 3.00 m/s, you
notice that one of the car’s tires has a small
hemispherical bump on its rim, as shown.
Example 2
a) Explain why the bump, from your viewpoint behind the car, executes simple harmonic motion.
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b) If the radii of the car’s tires are 0.300 m, what is the bump’s period of oscillation ?
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A simple pendulum exhibits periodic motion. It consists of a particle-
like bob of mass m suspended by a light
string of length L that is fixed at the upper end.
Lesson 4 : The Pendulum
Forces acting on bob
Tension in string
Gravitational force mg
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Ft = mg sin always acts opposite to the displacement
of the bob
Ft is a restoring force
Ft = -mg sin = md2s
dt2
Since s = L and L is constant,
d2dt2
= -g
Lsin for small values of
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d2x
dt2= -
k
mx
d2dt2
= -g
Lsin
same form
For small amplitudes ( < about 10o), the motion of a pendulum is close to simple
harmonic motion.
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Instead of x = A cos(t + ),
= max cos(t + )
Instead of = k
m,
= g
L
angular frequency
for a simple pendulum
T =2
= 2L
g
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= g
LT = 2
L
g
The period and frequency of a simple pendulum depend only on the length of the string and the acceleration due to gravity.
(Period and frequency are independent of mass.)
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Example 1
Christian Huygens (1629-1695), the greatest clockmaker in history, suggested that an
international unit of length could be defined as the length of a simple pendulum having a
period of exactly 1s.
a) How much shorter would our length unit be had his suggestion been followed ?
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b) What if Huygens had been born on another planet ? What would the value of g have to be on that planet such that the meter based on Huygen’s pendulum would have the same value as our meter ?
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Example 2
A simple pendulum has a mass of 0.250 kg and a length of 1.00 m. It is displaced
through an angle of 15.0o and released. What is the
a) maximum speed
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c) maximum restoring force ?
b) maximum angular acceleration
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Example 3
A simple pendulum is 5.00 m long.
a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 5.00 m/s2 ?
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b) What is its period if the elevator is accelerating downward at 5.00 m/s2 ?
c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 5.00 m/s2 ?
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The Physical Pendulum
A hanging object that oscillates about a fixed axis that does not pass through its center of mass and the object cannot be
approximated as a point mass.
Gravitational force produces a torque about
an axis through O.
= mgd sin
Since = I,
- mgd sin = Id2dt2
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- mgd sin = Id2dt2
negative sign because tends to decrease (restoring force)
If is small so that sin is almost ,
mgdd2dt2
= - ( )I
= - 2
simple harmonic motion equation
=mgd
I
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Since the period of a pendulum is
T =2
,
T = 2 I
mgd
Period of a Physical Pendulum
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A uniform rod of mass M and length L is pivoted about one
end and oscillates in a vertical plane. Find the period of
oscillation if the amplitude of the motion is small.
Example 4
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A rigid object suspended by a wire. When the object is twisted, the twisted wire exerts a restoring torque that is proportional to the
angular position.
Torsional Pendulum
= -
is called the torsion constant of the wire
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Since = Id2dt2
,
simple harmonic motion equation
d2dt2
I
= -
= I
T = 2 I
Period of a Torsional
Pendulum
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Example 5
A torsional pendulum is formed by taking a meter stick of mass 2.00 kg, and attaching to its center a wire. With its upper end clamped,
the vertical wire supports the stick as the stick turns in a horizontal plane. If the resulting period is 3.00 minutes, what is the torsion
constant for the wire ?