Transcript Lecture #2 x(t) = xm cos( t + )

Oscillations
x(t) = xm cos(t + )
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x(t)=xm cos(t+)
v(t)=- xm sin (t+ )
vm= xm ‘amplitude’
shifted by T/4 (900)
a(t)=- 2xm cos(t+
)
• am = 2xm ‘amplitude’
• shifted by 2T/4 (1800)
2 =- 22x
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d2x/dt
x   x
Example
• A body oscillates with SHM according to
x(t)= (6.0m) cos( 3t + /3)
• At t=2.0 s, what are (a) the displacement,
(b) the velocity,(c) the acceleration, (d) the
phase of the motion,(e) the frequency, (f)
the period ?
Solution
• x(t)= (6.0m) cos( 3t + /3)
• xm=6.0m, =3 rads/s,  =/3 rads
(constants!)
• (a) x(t=2) = 6cos(6+
/3)=6cos(600)=3.0m
• (b) v(t)= -(3)(6)sin(3t + /3)
v(t=2)=-18sin(6+/3)=1831/2/2 m/s
• (c) a(t) = - (3)2(6) cos( 3t + /3)
Solution cont’d
• x(t)= (6.0m) cos( 3t + /3)
• (d) phase = t+ = 3t + /3 = (19/3)
rads
• (e) = 2/T = 2 f ==> f = / 2 = 1.5 s1
• (f) T=1/f = 2/3 s
Force Law for SHM
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Newton’s second law F = m a
‘a’ non-zero ==> there is a force
F = ma = m(- 2x) = -m 2x = - k x
force  -displacement (in oppositedirection)
Hooke’s law for springs with k = m 2
SHM
d2x/dt2 = - 2x
or
F = -kx
a= - 2x plus F=ma ==> F= -k x where k=m 2
=(k/m)1/2
T= 2/ = 2(m/k)1/2
Example
• A small body of mass 0.12 kg is undergoing
SHM of amplitude 8.5 cm and period 0.20 s
• (a) what is maximum force?(b) if the
motion is due to a spring, what is k?
• What do we have to know?
• (1) F=ma (2) F=-kx
(3) a= - 2x
• or in other words x(t)=xm cos(t+)
Solution
• (a) Fmax = m amax = m|2xm |
(1)+(3)
• =2/T = 2/.2 = 10  rads/s
• Fmax = (.12 kg)(.085m)(10 s-1)2 =10. N
• (b) k= m2 =(.12kg)(10s-1)2
=1.2x102N/m
(2)+(3)
Uniform Circular Motion
P is projection of P`
x(t)=xm cos(t+)
x=rcos
Velocity
v=r 
vx=-vsin
vx(t)= - xm sin(t+)
a=v2/r=r2=xm 2
acceleration
ax(t)= -2 xm cos(t+)
Uniform Circular Motion
• Uniform circular motion
• r(t) = x(t) i + y(t) j
• x(t)= xm cos(t+), y(t)= xm sin(t+)
• r(t) is the sum of SHM along perpendicular
directions
Problem
• Two particles execute
SHM of the same
amplitude and
frequency along a
straight line.They pass
each other moving in
opposite directions
each time their
displacement is half
their amplitude. What
is the phase difference
between them?
Where are the particles on the
circle when they pass?
Solution
• x(t)=xm cos(t+)
• x(t) = xm /2 when
phase angle = ± 600
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phase difference= 1200
Energy in SHM
• As particle oscillates, its speed varies and
hence so does its kinetic energy K
• Where does the energy go when speed is
zero?
• If the oscillation is produced by a spring,
the spring is compressed or stretched to
some maximum amount when speed is zero
• Energy is in the form of potential energy U
Energy in SHM
• U(t) = (1/2) k x2 = (1/2)k xm2 cos2(t+)
• K(t) = (1/2) mv2 = (1/2) m2 xm2 sin2(t+
)
• using k = m 2 and cos2 + sin2 =1
• E(total) = U(t)+K(t) =(1/2) k xm2 (constant!)
=(1/2) m2 xm2