Maquinaria naval

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12 – Oscillations and waves

PHY166 Fall 2005

Oscillations or vibrations are periodic motions in physical systems (such as mass on a spring) under the influence of restoring forces. Waves are motions of distributed systems (such as string) that are periodic in both time and space.

Oscillations
Example: mass on a spring Restoring force:

m

F = −kx, k > 0 (stiffness constant)
Kineticenergy: mv 2 Ek = 2 Potential energy: kx 2 Ep = 2

x (0)

a + ω2x = 0

− General equation for all kinds of oscillating systems (oscillators)

In our case ω =

k m

ω is the angular frequency of oscillations

1

Note that Newton’s second law does not result in the case of oscillations to a motion with constant acceleration because the force is not constant and depends on thedisplacement. Solutions of Newton’s second law for oscillations in the general form

a + ω2x = 0
can be easily obtained with the calculus. It has a sinusoidal form

x = A cos(ωt + ϕ0 )

where A is the amplitude of oscillations, ω is the anguilar velocity and ϕ0 is a phase that depends on the initial conditions. Once x(t) is known, one obtains the acceleration: 2 2

a = −ω x = −ω A cos(ωt + ϕ 0)

the velocity v can be obtained with the calculus or, alternatively, from the energy conservation law. At the turning points of the motion where x = ±A the velocity is zero and the whole energy is potential energy kA2/2 . Thus the energy conservation can be written in the form
mv 2 kx 2 kA2 + = 2 2 2
¦

v

2

¥

¢

ω

+ x 2 = A2

£

¤

It follows then

v = ω A 2 − x 2 =ωA 1 − cos 2 (ωt + ϕ 0 ) = ωA sin 2 (ωt + ϕ0 ) = ±ωA sin(ωt + ϕ0 )
Thus, all together,

¡

 

− ωA sin(ωt + ϕ 0 )

x = A cos(ωt + ϕ 0 ) v = −ωA sin(ωt + ϕ 0 ) a = −ω 2 A cos(ωt + ϕ 0 )
− for all oscillators!

(Correct sign can be easily obtained with calculus) 2

§

Problem
Find the phase ϕ0 and the dependence x(t) for the oscillatory motion that starts at t = 0 in the state where(a) velocity is zero and the displacement is maximal; (b) velocity is zero and the displacement is minimal; (c) displacement is zero and velocity is positive; (d) displacement is zero and velocity is positive. Solution: (a) take the general solution x = A cos(ωt + ϕ 0 ) and plug t = 0 and x = A:

A = A cos(ϕ 0 ) x = A cos(ωt )

1 = cos(ϕ0 )

ϕ0 = 0

 

(b) take x = A cos(ωt + ϕ 0 ) andplug t = 0 and x = −A:

− A = A cos(ϕ 0 )

− 1 = cos(ϕ0 )

¡

x = A cos(ωt + π ) = − A cos(ωt )
(c) take x = A cos(ωt + ϕ 0 ) and plug t = 0 and x = 0:

¡
π

0 = A cos(ϕ0 )

0 = cos(ϕ0 )

ϕ0 = ±

¢

¢

We see that there are two solutions. To find the proper one consider the velocity at t = 0:

§ ¦ ¨

  

x = A cos ωt − x = A cos ωt +
 

2

  

Here wetake the upper sign in (c) and obtain

π

 © 

= µωA 2 Positive velocity corresponds to the lower sign, thus take the lower sign to obtain

v (t = 0) = −ωA sin(ϕ 0 ) = −ωA sin ±

π

= A sin(ωt )
3

2

= − A sin(ωt )

¤ £ ¥

 

ϕ0 = π

π
2

Plots for (a-d)

x
4

(a)
A=4
2

(c)

0

(d)
-2

(b)
-4

T
Frequency and period of oscillations ω Frequency: f= Period: 2π
T= 1 2π = f ω

t
4

Problem Mass m = 0.5 kg is attached to a spring with the stiffness constant k = 20 N/m on a horizontal frictionless table. The mass is pushed with the velocity v = 2 m/s in the
positive direction out of the equilibrium position. What is (a) the amplitude of harmonic oscillations; (b) the maximal acceleration; (c) full time dependence x(t)? (d) Time toachieve maximal displacement for the first time?

mv 2 kx 2 kA2 Solution: (a) Use the energy conservation in the form + = 2 2 2 In the initial state x = 0 and v is known, thus the amplitude A is
A2 = m 2 v k A= m 0.5 v= 2 = 0.316 m k 20
F k =− x m m

(b) From Newton’s second F=ma follows a =

The maximal acceleration corresponds to the maximal x, that is, to x=A. One obtains
amax = k k m k...
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