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MECHANICAL VIBRATIONS ME 65

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x1= aSin?1t and x2 = b Sin?2 t moving in same direction ,?1 is not equal to ?2, ... The resultant amplitude A2=(Acos?) 2 (Asin? )2. A = 9.42 ... – PowerPoint PPT presentation

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Title: MECHANICAL VIBRATIONS ME 65


1
MECHANICAL VIBRATIONS(ME 65)
  • Session 2
  • By
  • Dr. P.Dinesh
  • Sambhram Institute of Technology
  • Bangalore

2
In Session 2
  • Phenomenon of Beats
  • Harmonic motion in complex form
  • Addition of Harmonic motions
  • Analytical
  • Graphical

3
  • A simple harmonic motion can be expressed as
  • xA sin?t
  • A amplitude of vibration
  • ? circular frquency, rad/sec
  • Consider a vector of length equal to amplitude A
    and rotating at a speed of ? rad/sec. After time
    t it will be at an angle ?t wrt ref. axis (X
    axis)
  • Projection of this vector on Y axis gives the
    diplacement vector at time t.

4
Y
P
A
?t
X
5
  • x A sin?t , projection on Y axis disp vector
  • x ?A(sin?t 90º) , gives velocity vector
  • x ?2A (sin?t 180º) , gives accn. vector
  • All vectors rotate in CCW direction

6
Beats Phenomenon
  • x1 aSin?1t and x2 b Sin?2 t moving in same
    direction ,?1 is not equal to ?2,resultant motion
    is not harmonic
  • Phase difference between x1 and x2 keeps changing
  • Resultant ampl. is (ab) when in phase and (a-b)
    when out of phase.
  • Resultant amplitude continuously varies from
    (ab) a (a-b)
  • This is Beats occuring with a frequency of ?2- ?1

7
Harmonic motion in Complex form
  • If r is vector magnitude, ? the angle it makes
    with ref.
  • The diplacement vector xrei?t
  • The velocity vector (dx/dt) i?x
  • The acceleration vector (dx/dt)2 -?2x
  • Vel. Leads disp by 90º and accn. Vector leads
    vel. Vector by 90º

8
Addition of Harmonic motions
  • Two vectors rep.SHM can be added
  • Analytically
  • x1 A1Sin?t , x2 A2Sin(?tf)
  • x x1 x2 the resultant motion is SHM
  • x A Sin(?t?), A res. Ampl., ?Ang. Of Resul.
    with x1

9
  • Add the following vectors analytically
  • x1 4cos(?t 10º) and x2 6sin(?t 60º)
  • x x1x2 A sin(?t ?)
  • A sin(?t ?) 4cos(?t 10º) 6sin(?t 60º)
  • Expanding both sides
  • A(sin?tcos ?cos?tsin?)4cos?tcos10º -
    4sin?tsin10º 6sin?tcos60º 6cos?tsin60º

10
  • sin?t(A cos?) cos ?t(Asin?) sin?t(2.305)
    cos?t(9.135)
  • Therefore, A cos? 2.305
  • A sin? 9.135
  • The resultant amplitude A2(Acos?) 2 (Asin? )2
  • A 9.42
  • tan ? (Asin?/Acos?)(9.135/2.305) 76º
  • Or x 9.42sin(?t76º)

11
  • Graphical Method
  • Putting two vectors in same form
  • x24cos(?t10º) 4cos(?t10º90º)
  • 4sin(?t100º)
  • x1 6 sin(?t60º)
  • Draw ox the reference axis.Draw oa of length 6
    units(60mm) wrt ox at 60º.Draw ob of 4 unit
    length (40 mm) at 100º to ox.

12
  • Complete the vector diagram by drawing lines
    parallel to oa and ob to get point c. measure oc
    which will be equal to A 9.42 units and ? the
    angle of oc wrt ox will be 76º

13
Problems
  • a) Add the following harmonics analytically and
    check the solution graphically.
  • X1 3sin(?t30º), X2 4 cos(?t10º)
  • b) Split the harmonic motion x 10
    sin(?t30º)into two harmonic motion, one having a
    phase angle of zero and 46º.

14
  • c) A body is subjected to two harmonic motions
    x1 15sin(?t 30º) and x2 8cos(?t60º). What
    extra harmonic should be given to bring it to
    static equilibrium
  • d) A harmonic displacement is given by
  • x(t) 6 sin(20t60º), find a) frequency and
    period of motion b) max. displacement, velocity
    and acceleration.

15
Conclusion Session 2
  • What is Beat phenomenon
  • Methods of adding two harmonics analytically and
    graphically
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