FE Physics
Formulas & Explanation     Repetitive motions  (oscillations or vibrations)

Repetitive motion : a periodic motion about an equilibrium position

De distance x from the body to the equilibrium position is a function of time

Amplitude A is the maximal distance from the body to the central position

f = 1/T

Period  T       : duration of one oscillation (or vibration) in s

frequency f   :number of oscillations (or vibrations) per second  in Hz

Simple harmonic motion:

Displacement x(t) is a sine function of time

x(t) = A sin (2 π f t)         (when the body is at t= 0 s in a positive direction through the central position)

x(t)  displacement at   t = t (s)

A amplitude         (x en A : the same unit of length)

Example 1

A body is in a simple harmonic motion.  f = 20.0 Hz.    A = 5.0 cm

At  t= 0 s is the body in a positive direction through the central position . In other words   φ(0) = 0

Find the displacement at t = 0.018 s

x = A sin (2  π f t) = 5.0 sin (2 π x 20.0 x 0.018) = 5.0 sin (2.26) = (5.0) (0,77) = 3.85 cm

Remark:  set calculator to  RADIANS !!!

Example 2

At what times (t < 0.05 s) in the above example is the displacement  x = +3.0 cm ? x(t) = A sin (2 π f t)

3.0 = 5.0 sin (2 π 20 t)

3.0/5.0 = sin (2 π 20 t)

0.6 = sin (2 π 20 t)      ( sin-1 0.6 = 0.64 )

0.64 = 2 π 20 t

t  = 5.12 x 10-3 s

The second time  :

t= 0.025 – 5.12 x 10-3 = 0.0198 s =

19.8 x 10-3 s

Hooke’s Law

In a simple harmonic motion is also

F(t) = - k  x(t)          Hooke’s law

F(t)  the restoring force  is proportional the displacement  x(t)

F (t)  force in N

x (t)  displacement  in m

C      force constant  in N/m

Phase

phase φ : state of the motion after last passing through the  central position  (in positive direction)

φ = t/T

This formula as at t = 0 s  the phase= 0 is. This means at t = 0 s the body is moving in a positive direction through the central position.

Example

A body is in simple harmonic motion.  f = 5.0 Hz .

At  t=0 s the body moves trough the central position in positive direction    (  φ (0) = 0 )

At  t = 0.34 s  :   φ = 0.34/0.20 = 1.7 = 0.7

Maximum speed

If the body moves through the central position then the maximum speed is:

vmax = (2 π A)/ T

Then  φ = 0  (body moves through the central position in positive direction) or φ= 0.5

(body moves through the central position in negative direction)

Spring-mass system

T = 2 π √( m/C)           T in s ; m in kg ; C in N/m

Example

An object is attached to a spring and is vibrating   C = 10 N/m  T = 2.0 s

Calculate the mass of the object

2.0 = 2 π (m/10)

0.318 = (m/10)

0.1013 = m/10

m = 1.0 kg

Simple pendulum

T = 2 π /g        T in s ; in m ; g in m.s-2

Example

The duration of one oscillation is 1.0 s.

Find the length of the pendulum.

1.0 = 2 π √ℓ/9.81

1.0/(2 π) = √ ℓ/9.81

0.159 = √ ℓ/9.81

0.025 = ℓ/9.81

= 9.81 x 0.025 = 0.25 m

Energy in a horizontal vibrating  spring

Energy transformation always takes place between elastic potential energy and kinetic energy

The maximum of the kinetic energy  occurs when the object is passing through the central position.

The maximum in potential energy occurs in the extreme position

In an undamped vibration :  Emech =  Ek + Ep  = constant.

E mech = Ek + Ep = Ek,max = Ep,max

Emech = ½ m vmax2 = ½ m (2 π A/T)2 = ½ m (4 π2 A2 / T2) = (2 π2 A2)  /T2 = 2 m π2 A2 f2

Emech = ½ C A2      (Ep in extreme position)

T = 2 π √m/C     so     C = 4 π2 m/T2

This gives :

Emech = ½    (4 π2 m/T2) A2 = (2 m π2 A2) T2 =      2 m π2 A2 f2

Emech  in J

m   mass in kg

A    amplitude in m

f     frequency in Hz