•  Main points of today’s lecture: •  Springs and masses •  Simple harmonic motion of a spring:

•  Pendulum

Physics 231 Lecture 17

constants are and /

)cos(

)sin()cos(

02

02

0

0

AmktAa

tAvtAx

x

x

θωθωω

θωωθω

=

+−=

+−=+=

( )0max 2cos

2

ϑπθθ

π

+=

=

ftgLT

Concept Quiz

!me(s)

displacementx

5m

-5m2 4 6 8 10

whatistheamplitudeoftheharmonicoscilla!on?

a)  5mb)  10mc)  2sd)  4s

Concept Quiz

!me(s)

displacementx

5m

-5m2 4 6 8 10

whatistheperiodoftheharmonicoscilla!on?

a)  4sb)  2sc)  .25Hzd)  .5Hz

Concept Quiz

!me(s)

displacementx

5m

-5m2 4 6 8 10

whatisthefrequencyoftheharmonicoscilla!on?

a)  f=1/T=0.25/s=0.25Hzb)  f=1/T=0.5/s=0.5Hzc)  f=2π/T=π/2/s=1.57Hzd)  f=2π/T=0.25π/s=0.78Hz

Frequency vs Period

FrequencyvsPeriodTimetogofrom+Ato–Aandbackistheperiod.i.e.theperiodisthe!meittakestocompleteonefullcycle.Period=T

Frequency:f=1/T.eg,ifT=2secèf=½Hz

Iffislarge,periodissmall: f=1kHz=1000cyclespersecond T=1/(1000Hz)=0.001sec

Hz=1/s

time (s)

A

-A

Period and Frequency •  Sinusoidal functions are periodic

–  Adding 2π to the argument does not change the value of the function

•  The period is the time interval over which function repeats itself.

( ) ( )cos cos 2A Aθ π θ= +

θ (rad)

A

-A

( ) ( ) 2cos cos 2 cosA t A t A tπω π ω ωω

⎛ ⎞⎛ ⎞= + = +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

tθ ω=

T =2πω

Question 7.1 Harmonic Motion I a) 0

b) A/2

c) A

d) 2A

e) 4A

A mass on a spring in SHM has

amplitude A and period T. What is

the total distance traveled by the mass

after a time interval T?

Question 7.1 Harmonic Motion I a) 0

b) A/2

c) A

d) 2A

e) 4A

A mass on a spring in SHM has

amplitude A and period T. What is

the total distance traveled by the mass

after a time interval T?

In the time interval T (the period), the mass goes through one complete oscillation back to the starting point. The distance it covers is A + A + A + A (4A).

Graphical Representation of Motion •  Our equations: •  x=A cos(θ)=A cos(ωt+ θ0)

What is θ0?

•  a) 0 rad •  b) π/2 rad •  c) π rad •  d) 3π/2 rad

If θ0 the right time dependence then θ0=±2πn also will as well Recall ωT=2π

x

ω 3/ 4⎡⎣ ⎤⎦T +θ0 = ±2πn

ω 3/ 4⎡⎣ ⎤⎦ 2π /ω⎡⎣ ⎤⎦ +θ0 = ±2πn

cos ω 3T / 4⎡⎣ ⎤⎦ +θ0( ) = 1

θ0 = ±2πn− 3π / 2 = π / 2 ± 2πm

Example The motion of an object is described by the equation

x = (0.30 m) cos(πt/3),

where t is assumed to be in seconds. Find (a) the position, (b) velocity and (c) acceleration of the object at t = 0 and t = 0.60 s, (d) the amplitude of the motion, (e) the frequency of the motion, and (f) the period of the motion.

0general form: x Acos( t θ ); ω= + A 0.3m; = 0 θ 0=/ 3Hzω π= 1.05Hz, = f / (2 ) .17Hz; ω π= = ( ) T 1/ f 6s= =

( )0x Asin( t θ ) 0.3m sin(1.05t) (t is in sec.)ω= + =

0v - Asin( t θ )ω ω= + ( ) ( )1.05Hz 0.3m sin(1.05t)= −2a ω x 1.1x= − = −

at t 0 : x 0.3 m; = = v 0 ; = 2 a 0.33m / s= −

at t 0.6s : x 0.24 m; = = v 0.37m / s ; = − 2 a .26m / s= −

Graphical Representation of Motion

•  When x is a maximum or minimum, velocity is zero

•  When x is zero, the speed is a maximum

•  When x is a maximum in the positive direction, a is a maximum in the negative direction

•  x=A cos(θ)=A cos(ωt+ θ0) •  vx= -ωAsin (ωt+ θ0) •  ax= - ω2Acos(ωt+ θ0)= - ω2x

Conceptual question

•  A mass attached to a spring oscillates back and forth as indicated in the position vs. time plot below. At point P, the mass has

–  a. positive velocity and positive acceleration. –  b. positive velocity and negative acceleration. –  c. positive velocity and zero acceleration. –  d. negative velocity and positive acceleration. –  e. negative velocity and negative acceleration.

recallax = −k / m i x

Example

•  A 50 coil spring has a spring constant of 860 N/m. One end of a 50-coil spring is attached to a wall. An object of mass 45 kg is attached to the other end of the spring and the system is set in horizontal oscillation. What is the angular frequency of the motion? –  a) 2.39 Hz –  b) 4.37 Hz –  c) 5.21 Hz –  d) 6.85 Hz –  e) 9.22 Hz

km

ω =

•  If mass spring system was arranged vertically with the mass suspended from the 50 coil spring, how would the frequency change? –  a) it would be smaller because gravity subtracts from the spring

force at the bottom of its motion. –  b) it would be larger because gravity adds to the spring force at the

top of its motion. –  c) it would be exactly the same. Gravity only displaces the

equilibrium point so that the equilibrium length is greater.

860N / m45kg

= 4.37Hz=

Reading quiz