Physics 12 (Honours) Simple Harmonic Motion notes V 1.0

8/12/2019 Physics 12 (Honours) Simple Harmonic Motion notes V 1.0

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Physics 12 HonorsNotesSimple Harmonic Motion V1.1 (KWP) Page 1of 6

PHYSICS 12 HONOURS

[AP UNIT] Simple Harmonic Motion

For more information, refer to.

1. Physics: Principles with Applications (5th Edition): Chapter 11 Oscillations andWaves; Chapter 12 Sound2. Materials about AP Physics B & Cas well as other resources.

Hookes Law

Applied force acting on the spring

F

=

Fis applied force on the spring (N)kis spring constant (N/m)xis the extension of the spring (m)

The force which the spring exerts back is given by

F =This is ookes Law.

From Newtons Second Law, = =

Springs and energy

E =

=

x

Fs

x

a

F

x

For a spring, slope = k

(spring constant)

W = Area under graph

= = =

frictionless


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E = + = 12 + 12

At the ends, given that x = A(maximum displacement)then

= 1

2 + 1

2

= 0 + 12 Since total mechanical energy conserves,12 + 12 = 12 = =

When x=A then v=0

When x=0 (equilibrium) then v= vmax = 0 = = = 1

= 1

Question

An object of mass 0.10kg is attached to a spring with spring constant k = 200N/m. If the

object oscillates between x = 0.05m and 0.05 m, calculate

(a) ETOTAL ; (b) Max.| v | ; (c) aMAX

x

v

frictionless

Amplitude

E = + Conservationof energy


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(a)Since total mechanical energy conserves, we could calculate E totalat any point of the

motion. The simplest way is to calculate Etotalat x=A, at which v =0, which gives:

E = 12 + 12 2 = 12 (200 )0.05 + 0 = 0 . 2 5 (b) Maximum v occurs when x = 0; hence,

| | = =0.05m 200/0.10 = 2.2 m/s(c) Maximum a occurs when x = A; hence,

S.H.M. as sinusoidal waves

12 14

34

T: Period;

t: time.

x

tT

A

-A

= = ( 2 ) (2 )2 = 2 2 2 = 4 2

v

t

2A 2A T =

= 2 2

= 2

T

a

t

4A 4A

= = 200 /0.10 0.05=100 /x is opposite of a

dydx derivative of y with respect to of x(Calculus)


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From the x and a graphs,

But from the spring equation,

So

This is the relationship between period of oscillation of a mass

on a spring, the mass and the spring constant.

Question (Half of the test in one big question!)

A mass of 4kg is attached to a horizontal spring with k = 100 N/m. The object is displaced10cm and released. Assume a frictionless surface. Determinea) The amplitude

b) The periodc) The frequencyd) The total mechanical energye) The maximum velocityf) The maximum accelerationg) The velocity and acceleration when the object is 5cm from the equilibrium position.

a) Amplitude

b) Period

c) Frequency

d) Total mechanical energy

= compare with =

= 4 = 4 = = 4

= 2

Angular velocity

(rad/s)

= 2 =2

= 10 = 0.1

= 2 = 2 4100 = 1.26

= 1 = 1 = 0.79 = 12 = 12 (100 ) 0.1 =0.5

= A (max. disp.)(v = 0)

= 4 2

= 2


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e) Maximum velocity

f) Maximum acceleration

g) The velocity and acceleration when the object is 5cm from the equilibrium position

Comparing SHM to circular motion

, where vois maximum velocity and since

But comparing this to shown earlier

= ( ) = (100 /4 ) 0.1 = 2.5 / = ( )

0.5 = 12

4 = 0.25 => = 0.5 /

=

= 100 /

4 0.1

0.05

= 0.4 /

= ( ) = 100 4 0.05 =1.25 /

=

=

= =

=2

= 2 Or

12 = 12 = =

2 = 2 = = 2

= 2

= 12 = 12 (Conservation of Mechanical Energy)x = A (v = 0) x = 0 (v = )


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= 2 This is the relationship between the period, mass and spring constant of a mass attached to aspring in Single Harmonic Motion.

Natural frequency, f0, and Resonance

A mass oscillating on a spring does so with a specific period given by . That

spring has a corresponding frequency called natural or resonant) frequency, f0:

All material and structure have a natural frequency. In fact that have more than one naturalfrequency.

If an external forceacting on the structure has a frequency close or equal to f0this causes thevibration of the structure to be amplified. This is called resonance. To prevent catastrophicdamage, the frequencies in contact with the structure must be different from f0.

Music

For extra information on the topic of music, sound and waves, read the following sections in

your textbookPhysics: Principles with Applications (5th Edition)

11.7-11.12 on waves (esp. 11.11, 11.12)

12.1-12.7 on sound (esp. 12.5, 12.6, 12.7)

Test Date

April 7th, 2014 (Monday), Block C (for Honours Class).

= 2

= 1 = 12

fundamental f0

vibrating string

harmonics or overtones

Physics 12 (Honours) Simple Harmonic Motion notes V 1.0

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