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Northa Northa
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7 years ago
A 1.5-kg mass attached to spring with a force constant of 20.0 N/m oscillates on a horizontal, frictionless track.  At t= 0, the mass is released from rest at x= 10.0 cm.  (That is, the spring is stretched by 10.00 cm.)
(a) Determine the frequency of the oscillations.
(b) Determine the maximum speed of the mass.  Where does the maximum speed occur?
(c) Determine the maximum acceleration of the mass.  Where does the maximum acceleration occur?
(d) Determine the total energy of the oscillating system.
(e) Express the displacement as a function of time.
Textbook 
Physics

Physics


Edition: 4th
Author:
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Raj.SRaj.S
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4 years ago
woah
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#3
2 years ago
Given that 
    forceconstant   k   =   20.0N/m,   maximum displacement(amplitude)   A   =   10.0 cm  =   0.10    m
   mass   m   =   1.5kg
   a.   angularfrequency   ω   =   2πf   =   √(k / m)   =   √ ( 20 /1.5)   =   3.65 rad/s
         =>   frequency   f   =   3.65/ 2π   =   3.65 /6.28   =   0.581   /s
   b.   Maximumspeed   vmax   =   ω* A   =   3.65 *0.10   =   0.365 m/s
      maximum speed occurs asthe mass passes through the equilibrium position.
   c.   Maximumacceleration   amax   =   -ω2 * A   =   -3.652 * 0.10   =  -1.33   m/s2
         maximumacceleration accurs at maximum displacement from equilibrium, i.e.extreme position. -ve sign indicates that acceleration is directedtowards the equilibrium position.
   d.   Totalenergy   E   =   (1/2)* m * ω2 * A2
            E   =   0.5* 1.5 * 3.652 * 0.12
            E   =   0.0999   J   ˜   0.1   J
   e.   equation of SHM is
         displacement   x   =   A* sin ( ω t)
   Substitutingvalues   x   =   0.10* sin (3.65 t)
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