Energy in SHM & The Simple Pendulum Energy Considerations in SHM - - PDF document

▶

Feb 19, 2023 334 likes •450 views

Energy in SHM & The Simple Pendulum Energy Considerations in SHM The Simple Pendulum Homework 1 Kinetic and Potential Energy in SHM Potential energy U = 1 2 kx 2 = 1 2 kA 2 cos 2 ( t + ) U max = 1 2 kA 2 Kinetic energy

SLIDE 1

Energy in SHM & The Simple Pendulum

Energy Considerations in SHM
The Simple Pendulum
Homework

SLIDE 2

Kinetic and Potential Energy in SHM

Potential energy

U = 1 2kx2 = 1 2kA2 cos2 (ωt + φ) Umax = 1 2kA2

Kinetic energy

K = 1 2mv2 = 1 2mω2A2 sin2 (ωt + φ) K = 1 2kA2 sin2 (ωt + φ) Kmax = 1 2kA2

SLIDE 3

Kinetic and Potential Energy in SHM (cont’d)

SLIDE 4

Energy Transfer in SHM

SLIDE 5

Total Mechanical Energy in SHM E = K + U E = 1 2kA2 sin2 (ωt + φ) + 1 2kA2 cos2 (ωt + φ) E = 1 2kA2 E = K + U = 1 2mv2 + 1 2kx2 = 1 2kA2 v = ±

m

A2 − x2
= ±ω
A2 − x2

SLIDE 6

Example A block whose mass is 0.680 kg is fastened to a spring whose spring constant is 65.0 N/m. The block is pulled a distance x = 0.110 m from its equilibrium position at x = 0 on a friction- less surface and released from rest at t = 0. (a) What is the mechanical energy of the system? (b) What are the potential energy and kinetic energy of the system when the block is at x = (1/2)xmax?

SLIDE 7

Example A block whose mass is 0.680 kg is fastened to a spring whose spring constant is 65.0 N/m. The block is pulled a distance x = 0.110 m from its equilibrium position at x = 0 on a friction- less surface and released from rest at t = 0. (a) What is the mechanical energy of the system? (b) What are the potential energy and kinetic energy of the system when the block is at x = (1/2)xmax? (a) E = K + U = 1

2mv2 + 1 2kx2

E = 0 + 1

2(65.0 N/m)(0.110 m)2 = 0.393 J

(b) U=1

2kx2 = 1 2k

 1

2xmax

 2 = 1

4

 1

2kx2 max

 =1

4E

U = 1

4 (0.393 J) = 0.0983 J

K = E − U = 0.393 J − 0.0983 J = 0.295 J

SLIDE 8

The Simple Pendulum

SLIDE 9

The Simple Pendulum (cont’d) τ = Iα −mgL sin θ = Id2θ dt2 For small oscillations sin θ ≃ θ −mgLθ = mL2d2θ dt2 d2θ dt2 + g Lθ = 0 θ = θmax cos (ωt + φ) ω =

L T = 2π ω = 2π

g

SLIDE 10

Homework Set 23 - Due Mon. Nov. 8

Read Sections 12.3 - 12.4
Answer Questions 12.8, 12.9, 12.10 & 12.11
Do Problems 12.15, 12.16, 12.19, 12.21, 12.22,