CH 10: SIMPLE HARMONIC MOTION & ELASTICITY (NEW)lightcat-files.s3.amazonaws.com/packets/admin_physics-3-physics... · ch 10: simple harmonic motion & elasticity (new) Page 8 PRACTICE

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PHYSICS - CUTNELL 11E

CH 10: SIMPLE HARMONIC MOTION & ELASTICITY (NEW)

CONCEPT: Hooke’s Law & Springs ● When you push/pull against a spring (FA), spring pushes back in the _________ direction. (Action-Reaction!)

| Fs | = | FA | = _____

Ex. 1: You push on a spring with a force of 120N. The spring constant k is 20. How much does it compress?

● x = D______________

- Relaxed position ______ (x = ___)

- NOT the spring’s length (x = _______)

● k = spring’s force constant

- Measures how ________ the spring is.

- Higher k _________ to deform

Ex. 2: How much force is required to pull a spring of length 10m out to 16m, if the spring constant k is 40N/m?

- Ex. 1: x = k = F =

- Ex. 2: x = k = F =

- Units of k: ______ ● FS = R___________ force, always opposes deformation

EXAMPLE 3: A 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring’s force constant? (b) How much force is needed to compress it to 0.7 m?



Page 2

PRACTICE: A 1.0 m-long spring is laid horizontally with one of its ends fixed. When you pull on it with 50 N, it stretches to 1.2 m. (a) What is the spring’s force constant? (b) How much force is needed to compress it to 0.7 m?



Page 3

CONCEPT: Spring Forces

● If you attach a mass to a spring (mass-spring system) and release, the ____________ force pulls it back to equilibrium.

- The “m” always refers to the mass of the _________ (springs are always massless!)

- Compressed: ΣF = ma

→ a = _______

- Released: ΣF = ma

→ a = _______

EXAMPLE 1: A 0.60-kg block attached to a spring with force constant 15 N/m. The block is

released from rest when the spring is stretched 0.2 m to the right. At the instant the block

is released, find (a) the force on the block and (b) its acceleration, assuming “right” is

positive.

PRACTICE: You push a 3-kg mass against a spring and release it from rest. Its maximum acceleration is 10m/s2 when pushed back 0.5m. What is the (a) spring constant and (b) restoring force at this point?

m

m

m



Page 4

CONCEPT: Intro to Simple Harmonic Motion

● The most common type of Simple Harmonic Motion (aka Oscillation) is the mass-spring system.

(EQ) ● Amplitude → A:

- ____________ displacement, |x|

- always the _________________.

● Period → T [seconds/cycle]

- Time for one ____________ cycle.

● Frequency → f = 1/T [cycles/second]

● Angular frequency → ω [rad/second]

- ω = _______ = _______

x = ______ = ______

v = ______

F = ______

a = ______

x = ______

v = ______

F = ______

a = ______

x = ______ = ______

v = ______

F = ______

a = ______

EXAMPLE 1: A mass on a spring is pulled 1m away from its equilibrium

position, then released from rest. The mass takes 2s to reach maximum

displacement on the other side. Calculate the (a) amplitude, (b) period, (c)

angular frequency of the motion.



Page 5

PRACTICE: A mass-spring system with an angular frequency ω = 8π rad/s oscillates back and forth. (a) Assuming it starts

from rest, how much time passes before the mass has a speed of 0 again? (b) How many full cycles does the system

complete in 60s?

Mass-Spring SHM Equations

|FS| = |FA| = kx

a = −k

mx

ω = 2π𝑓 =2π

T= √

k

m

𝑁 [cycles] =𝑡 [time]

𝑇 [Period]= 𝑡 ∗ 𝑓



Page 6

CONCEPT: Equations for Simple Harmonic Motion

● In Simple Harmonic Motion, acceleration NOT constant → kinematic equations?

Old Equations → x (position)

Fs = −k x → Fmax = _____

a = −k x

m → amax = _____

New Equations → t (time)

x(t) = + A cos(ωt) → xmax = _____

v(t) = − Aω sin(ωt) → vmax = _____

a(t) = −Aω2cos(ωt) → amax = _____

- Calculator must be in _________.

● Combining amax(x) and amax(t) → 𝜔 = 2𝜋𝑓 =2𝜋

𝑇= _______

EXAMPLE 1: A 4-kg mass is attached to a spring where k = 200[N/m]. The mass is pulled

2m and released from rest. Find the (a) angular frequency, (b) velocity 0.5s after release,

(c) acceleration when x = 0.5m, and (d) the period of oscillation.



Page 7

PRACTICE: A 4-kg mass on a spring is released 5 m away from equilibrium position and takes 1.5 s to reach its equilibrium

position. (a) Find the spring’s force constant. (b) Find the object’s max speed.

EXAMPLE: A 4-kg mass is attached to a horizontal spring and oscillates at 2 Hz. If mass is moving with 10 m/s when it

crosses its equilibrium position, (a) how long does it take to get from equilibrium to its max distance? Find the (b) amplitude

and (c) maximum acceleration.

-A 0 +A


|FS| = |FA| = kx → Fmax = ±kA

a = −k

mx → amax = ±

k

mA

x(t) = + A cos(ωt) → xmax = ±A

v(t) = −Aω sin(ωt) → vmax = ±Aω

a(t) = −Aω2 cos(ωt) → amax = ±Aω2

ω = 2π𝑓 =2π

T= √

k

m



-A 0 +A



a = −k

mx → amax = ±

k

mA




ω = 2π𝑓 =2π

T= √

k

m





Page 8

PRACTICE: What is the equation for the position of a mass moving on the end of a spring which is stretched 8.8cm from

equilibrium and then released from rest, and whose period is 0.66s? What will be the object’s position after 1.4s?

EXAMPLE: The velocity of a particle on a spring is given by the equation v(t) = -6.00 sin (3π t). Determine the (a) frequency

of motion, (b) amplitude, and (c) velocity of the particle at t = 0.5s.

-A 0 +A



a = −k

mx → amax = ±

k

mA




ω = 2π𝑓 =2π

T= √

k

m



-A 0 +A



a = −k

mx → amax = ±

k

mA




ω = 2π𝑓 =2π

T= √

k

m





Page 9

CONCEPT: Energy in Simple Harmonic Motion

● At any point of SHM, the mass-spring system may have 2 types of Energy: ____________ + _____________.

- Wnc = _____, so the Mechanical Energy (M.E.) is ________________.

(EQ)

● Energy Conservation → always compare energies at 2 special points: Any other Point:

x = ______

Elastic Energy (UP) = 1

2𝑘𝑥2 = ______

Kinetic Energy (KP) = 1

2𝑚𝑣2 = ______

Total M.E. = _______________

Amplitude:

x = ______

Elastic Energy (UA) = 1

2𝑘𝑥2 = ______

Kinetic Energy (KA) = 1

2𝑚𝑣2 = ______

Total M.E. = _______________

Equilibrium:

x = ______

Elastic Energy (U0) = 1

2𝑘𝑥2 = ______

Kinetic Energy (K0) = 1

2𝑚𝑣2 = ______

Total M.E. = _______________

● Comparing all these energies at different points:

𝑈𝐴 = 𝐾0 = 𝑈𝑃 + 𝐾𝑃

_________ = __________ = __________ + __________ → 𝑣(𝑥) = __________________

(Energy Conservation for Springs)

EXAMPLE 1: A 5 kg mass oscillates on a horizontal spring with k = 30[N/m] and an amplitude of 0.4 m. Find its (a) max

speed, (b) speed when it is at -0.2 m, and (c) the total mechanical energy of the system.

m



Page 10

EXAMPLE: A 0.25-kg mass oscillates on a spring with a period of 3.2s. At x=0.4m, it is observed to have a speed of 5m/s. What is the

system’s (a) Amplitude and (b) total mechanical energy?

PRACTICE: A block of mass 0.300 kg is attached to a spring. At x = 0.240 m, its acceleration is ax = -12.0 m/s2 and its velocity is vx =

4.00 m/s. What are the system’s (a) force constant k and (b) amplitude of motion?

-A 0 +A



a = −k

mx → amax = ±

k

mA




ω = 2π𝑓 =2π

T= √

k

m



𝑀. 𝐸. =1

2𝑘𝐴2 =

1

2𝑚𝑣𝑚𝑎𝑥

2 =1

2𝑘𝑥𝑝

2 +1

2𝑚𝑣𝑝

2

𝑣(𝑥) = 𝜔√𝐴2 − 𝑥2

-A 0 +A



a = −k

mx → amax = ±

k

mA




ω = 2π𝑓 =2π

T= √

k

m



𝑀. 𝐸. =1

2𝑘𝐴2 =

1


2 =1

2𝑘𝑥𝑝

2 +1

2𝑚𝑣𝑝

2

𝑣(𝑥) = 𝜔√𝐴2 − 𝑥2



Page 11

EXAMPLE: You increase the amplitude of oscillation of a mass vibrating on a spring. Which statements are correct?

(a) Period of oscillation increases (b) Maximum acceleration increases (c) Maximum speed increases

(c) Max Kinetic Energy increases (d) Max Potential Energy increases (e) Max Total Energy increases

-A 0 +A



a = −k

mx → amax = ±

k

mA




ω = 2π𝑓 =2π

T= √

k

m



𝑀. 𝐸. =1

2𝑘𝐴2 =

1


2 =1

2𝑘𝑥𝑝

2 +1

2𝑚𝑣𝑝

2

𝑣(𝑥) = 𝜔√𝐴2 − 𝑥2



Page 12

CONCEPT: Vertical Oscillations

● Vertical mass-spring systems very similar to Horizontal, except:

- Horizontal → Equilibrium at relaxed position (x = 0)

- Vertical → Equilibrium where forces ____________.

EXAMPLE: You hang a 0.5m spring from the ceiling. When you attach a

5kg mass to it, it stretches by 0.2m. You pull the mass-spring system

down an additional 0.3m and release. Find (a) the spring constant.

(b) At its maximum height, how far from the ceiling is the block?

EQ



Page 13

PRACTICE: A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is then pulled down 6.0 cm and released. If the ball makes 30 oscillations in 20 s, what are its (a) mass and (b) maximum speed? EXAMPLE: An elastic cord is 65 cm long when a weight of 75 N hangs from it but is 85 cm long when a weight of 180 N hangs from it.

What is the “spring” constant k of this elastic cord?

PRACTICE: A chair of mass 30 kg on top of a spring oscillates with a period of 2s. (a) Find the spring’s force constant. You place an

object on top of the chair, and it now oscillates with a period of 3s. (b) Find the object’s mass.

-A 0 +A


|F�| = |F�| = kx → Fmax = ±kA

a = −�

�x → amax = ±

�

�A



a(t) = −Aω� cos(ωt) → amax = ±Aω�

ω = 2π� =2π

T= �

k

m

� [cycles] =� [time]

� [Period]= � ∗ �

�. �. =1

2�� =

1

2��

� =1

2��

� +1

2��

�

�(�) = �� − ��



Page 14

CONCEPT: Simple Pendulums ● Just like mass-spring systems, pendulums also display Simple Harmonic Motion.

Mass-Spring

F = ______ a = ______

� = 2�� =��

�= ______

Pendulum

F = _______ ≈ ______ a = _______ ≈ ______

� = 2�� =��

�= ______

- Make sure θ and calculator is in RADIANS! ● For SHM, restoring force must be proportional to deformation/distance.

- For small angles, _______ ≈ _______. → Restoring Force: __________

Example 1: You pull a 0.250m long pendulum with a hanging mass of 4kg to the side by 3.50° and release. Find the (a) restoring force, (b) period of oscillation, (c) time it takes for the mass to reach its maximum speed.

m

m

(EQ)

m



Page 15

PRACTICE: A pendulum makes 120 complete oscillations in 3.00 minutes. Find (a) the period of oscillation and (b) the length of the pendulum.

EXAMPLE: After landing on an unfamiliar planet, an astronaut constructs a simple pendulum of length 3m and mass 4kg. The astronaut releases the pendulum from 10 degrees with the vertical, and clocks one full cycle at 2s. Calculate the acceleration due to gravity at the surface of this planet.

Pendulum SHM Equations

|F�| = −mgθ

a = −gθ = −�

��

ω = 2π� =2π

T= �

�

�


� [Period]= � ∗ �


|F�| = −mgθ

a = −gθ = −�

��

ω = 2π� =2π

T= �

�

�


� [Period]= � ∗ �



Page 16

CONCEPT: Pendulum SHM Equations

-A 0 +A

● If asked for � given t → �(�) = ��(��)

EXAMPLE: A 500-g mass hangs from a 40-cm-long string. The object has a speed of 0.25m/s as it passes through its lowest point. What maximum angle (in degrees) does the pendulum reach?

Mass-Springs Pendulums

x = deformation from EQ At any point:

x = ________ ≈ ________

xmax = A

vmax = Aω

amax = Aω2

xmax = _______ = _______

vmax = Aω = ___________

amax = Aω2 = ___________

ω =2πf = 2π/T=��

� ω =2πf = 2π/T= �

�

�

m



Page 17

PRACTICE: A simple pendulum is 0.30m long. At t = 0, it is released from rest, starting at an angle of 13°. What will be the

angular position (in degrees) of the pendulum at t = 0.35s?

EXAMPLE: A 100g mass on a 1.0m-long string is pulled 7.0° to one side and released. How long does it take to reach 4.0° on the opposite side?


|F�| = |F�| = −mgθ

a = −gθ = −�

��

ω = 2π� =2π

T= �

�

�


� [Period]= � ∗ �

→ xmax = A = Lθ

�(�) = ��(��) → vmax = Aω = Lθmaxω

→ amax = Aω2= Lθmaxω2


|F�| = |F�| = −mgθ

a = −gθ = −�

��

ω = 2π� =2π

T= �

�

�


� [Period]= � ∗ �

→ xmax = A = Lθ

�(�) = ��(��) → vmax = Aω = Lθmaxω




Page 18

CONCEPT: Energy in Simple Pendulums

● Just like mass-spring systems, energy in pendulums → 2 types: ___________ and _________________.

● For any θ, height h = ____________

Amplitude:

θ = ______

Grav. Potential = ��ℎ = 0 / max

Kinetic Energy = �

�� = 0 / max

Total M.E. = _________

Equilibrium:

θ = ______

Grav. Potential = ��ℎ = 0 / max


�� = 0 / max

Total M.E. = _________

Any other Point:

θ = ______

Grav. Potential = ��ℎ


��

Total M.E. = ________+_______

M.E. = __________ = __________ = ___________+__________

(Energy Conservation for Pendulums)

EXAMPLE: A mass m is attached to a pendulum of length L. It is pulled up an angle θ and released. Using energy

conservation, derive an expression for the maximum speed this mass experiences.

m



Page 19

EXAMPLE: A mass of 0.400 kg hangs from a 2m pendulum. At the moment when it makes a 5° with the vertical, it has a

speed of 1.5 m/s. What is the maximum height the pendulum will reach?

PRACTICE: A mass swinging at the end of a pendulum has a speed of 1.32m/s at the bottom of its swing. At the top of its swing, it makes a 9° with the vertical. What is the length of the pendulum?


|F�| = |F�| = −mgθ

a = −gθ = −�

��

ω = 2π� =2π

T= �

�

�


� [Period]= � ∗ �

→ xmax = A = Lθ

�(�) = ��(��) → vmax = Aω = Lθmaxω



|F�| = −mgθ

a = −gθ = −�

��

ω = 2π� =2π

T= �

�

�


� [Period]= � ∗ �

�(�) = ��(��) A = Lθ

�� = �2��(1 − cos ��) → vmax = Aω

ℎ = �(1 − ��)

�. �. = ��ℎ�� =1

2��

� = ��ℎ� +1

2��

�



Page 20

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CH 10: SIMPLE HARMONIC MOTION & ELASTICITY (NEW)lightcat-files.s3.amazonaws.com/packets/admin_physics-3-physics... · ch 10: simple harmonic motion & elasticity (new) Page 8 PRACTICE