Example#1-SHM. The motion of a 0.25kg block around its equilibrium position in a spring-block
system is described by the function y = -2 sin(4t+π/2)[m]. Find:
a) The amplitude of oscillations.
From the function one gets A=2[m]
b) The angular frequency of oscillations.
From the function one gets ω = 4[r/s]
c) The natural frequency and period of oscillations.
f = ω/ 2π =4/6028 =0.64[Hz]
T=2π/ω=6.28/4=1.57[s]
d) The spring constant .
So,
ω2 = k / m ;
k = ω2 m = 42*0.25 =16*0.25 = 4[N/m]
e) The first and second time (after t=0s) when the spring is at its maximum extension.
Procedure : Whenever there is a question about time one must find first the corresponding phase at
that time ϕ = ωt + φ o. Next one should start by drawing the phasor at t=0s and its position at the phase
value found. Finally, one uses the relation between the phase and time to find the time.
In this problem, at t=0s y = -2 sin(π/2) = -2 = 2sin(-π/2) which shows that at t = 0s ϕ = φ o = - π/2.
One follows by drawing the phasor at t=0s. As the problem asks for maximum extension after t=0s
( y1max = +/- 2[m]), it is clear that this condition is realised for the first time (t1) when the phasor is
vertically up (φ1 = π/2) and for the second time (t2) when it is anew vertically down (φ2 = 3π/2).
Before finding the phase, one converts
the function as y = 2 sin(4t - π/2) and
after this notes φ = 4t - π/2 . Next:
φ1 = + π/2 = 4t1 - π/2 gives t1 = 0.785s
t=0s; φ=ϕo=-π/2
t1; φ1= +π/2
t2; φ2= +3π/2
φ2 = + 3π/2 = 4t2 - π/2 gives t2 = 1.57s
e) The displacement as a "cos" function. As sin(x+π/2) = cosx , in our case
sin(4t+π/2) = cos4t
and we may write the SHM function as y = -2 cos(4t)[m] or y = 2 cos(4t+π)[m] (as cos(x+π) = - cosx)
f) The block velocity v = dy/dt = d/dt (-2 sin(4t+π/2) = (-2)(4)cos(4t+π/2) = -8cos(4t+π/2)[m/s]
i) The maximum magnitude of spring force. Fel = -ky and Fmax= abs(kymax) = kA = 4*2 = 8N
g) The mechanical energy of system , its maximum kinetic energy and maximum speed of block.
E=K+U= Umax . So, E = Umax= 0.5kA2 = 0.5*4*22 = 8J , Also, E = Kmax . So Kmax = 8J
As Kmax = 1/2mvmax2 it comes out that vmax=[2*8/0.25]1/2 = 8m/s
Or straight away from velocity expression at "f" one gets speedmax= max[abs(v)] = 8m/s and next,
finds out Kmax and E .
Example#2-SHM
Given the function governing the motion of a simple pendulum with mass 0.5kg is
  0.5 cos(0.5 * t   / 2) [rad], find: a) The period of oscillations;
b) The length of pendulum;
c) the maximum angle reached in degree;
d) the position and linear velocity of bob at t = 0
e) kinetic energy of bob when the angle is θ = 0.2rads ;
f) total mechanic energy of pendulum
a) ω=0.5π . So, T=2π/ω=2π/0.5π=4sec.
b) ω2= g/L. So, L = g/ ω2= 9.8/(0.5*3.14)2= 3.98[m]
θ
L
s
c) From function θmax= 0.5[r]= 0.5[r] *(180o/3.14[r])= 26.66o
h
d) At t= 0s θ(0) =0.5cos(0+π/2) = 0.5cos(π/2)= 0[r] .
The linear velocity is calculated from the displacement . So,
one refers to the translational displacement from lowest
level, noted by letter "s" and counted as "+" on the rightside.
As s = L*θ one gets
v(t) = ds/dt = d(L*θ) / dt = Ldθ/dt = 0.5*(-0.5π)sin(0.5πt+π/2) = -3.12sin(0.5πt+π/2) (**) ,
At t=0 one gets v(0) = -3.12sin(π/2)= -3.12[m/s] which shows that the bob starts the motion right side.
e) K(θ=0.2r) = m*v2(θ=0.2r)/2 =0.25* v2(θ=0.2r). So, one has to find v-value when θ = 0.2r. For this,
one finds out the value of phase φ = 0.5πt+π/2 that corresponds to θ = 0.2r and then calculates (**) for
this phase value. Starting by 0.2 = 0.5cosφ or cosφ = 0.2/0.5 = 0.4, one gets φ = arscos0.4 = +/- 1.16[r]
Next, one uses this phase at (**) and gets K(θ=0.2r) = 0.25*(-3.12sin(+/-1.16))2= 2.04 Joules
f) Two ways f.1) E = Kmax = K(θ=0r) = (0.5/2)*v2max = 0.25 *(-3.12)2 = 2.4 Joules
or f.2) E =Umax= U(θmax) = mghmax = mgL(1- cosθmax) = 0.5*9.8*3.98(1- cos0.5r) = 2.4J.
Note that hmax = L - Lcosθmax = L(1- cosθmax)
Example#3-DHM. For a given damped oscillator with m=250g, k=85N/m, b=70g/s find: a) The period
of oscillations. b) How long does it take for the amplitude to drop to half of its initial value. c) How
long does it take for the mechanical energy to drop to one half of its initial value. d) What is the ratio of
the amplitude after 20 cycles compared to initial value.
a) ωo2= k/m= 85/0.25=340[r/s]2 and ω'=[ ωo2-(b/2m)2]1/2
ω' = [ 340-(70/250)2]1/2 = 18.43 r/s and T' =2π/ω'=0.34s
b) A=Ao/2=Aoe-(b/2m)*T1/2
e 0.14*T1/2 = 2
and
0.5= e-(70/2*250)*T1/2
0.14*T1/2 = ln2 =0.693
So, T1/2= 0.693/0.14 = 4.95[s]
c) E1 = 0.5kA12 = 0.5Eo = 0.5(0.5kAo2 ) = 0.25k Ao2 . So, A12 = 0.5Ao2 and A1 = 0.707Ao = Aoe-0.14*t
So, e 0.14*t = 1 / 0.707 = 1.414 and 0.14*t = ln1.414 = 0.346 and t = 0.346 / 0.14 = 2.475s
d) t = 20*0.34s = 6.8 s. So, A(6.8s) = Aoe-0.14*6.8 and
A(6.8s) / Ao
=
e -0.14*6.8 = 0.386