Oscillations

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Oscillations
x(t) xm cos(?t ? )

• 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)
• d2x/dt2 - ?2x

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Example

• A body oscillates with SHM according to x(t)
  (6.0m) cos( 3?t ?/3)
• At t2.0 s, what are (a) the displacement, (b)
  the velocity,(c) the acceleration, (d) the phase
  of the motion,(e) the frequency, (f) the period ?

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Solution

• x(t) (6.0m) cos( 3?t ?/3)
• xm6.0m, ?3? rads/s, ? ?/3 rads
  (constants!)
• (a) x(t2) 6cos(6? ?/3)6cos(600)3.0m
• (b) v(t) -(3?)(6)sin(3?t ?/3) v(t2)-18?sin
  (6??/3)-18?31/2/2 m/s
• (c) a(t) - (3?)2(6) cos( 3?t ?/3) a(t2)
  -54 ?2cos(600) -27 ?2m/s2

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Solution contd

• x(t) (6.0m) cos( 3?t ?/3)
• (d) phase ?t? 3?t ?/3 (19/3)? rads
• (e) ? 2?/T 2? f gt f ?/ 2? 1.5 s-1
• (f) T1/f 2/3 s

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Force Law for SHM

• Newtons second law F m a
• a non-zero gt there is a force
• F ma m(- ?2x) -m ?2x - k x
• force ? -displacement (in oppositedirection)
• Hookes law for springs with k m ?2
• SHM d2x/dt2 - ?2x
• or F - k x

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a - ?2x plus Fma gt F -k x where km
?2 ?(k/m)1/2 T 2?/? 2?(m/k)1/2
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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) Fma (2) F-kx (3) a - ?2x
• or in other words x(t)xm cos(?t?)

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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)

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Uniform Circular Motion
P is projection of P
x(t)xm cos(?t?)
xrcos?
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Velocity
vr ?
vx-vsin?
vx(t) -? xm sin(?t?)
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av2/rr?2xm ?2
acceleration
ax(t) -?2 xm cos(?t?)
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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

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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?
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Solution

• x(t)xm cos(?t?)
• x(t) xm /2 when phase angle 600

?
?
phase difference 1200
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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

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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

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