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Energy in Oscillator

Table of Contents
Introduction
Problems
Answers

      
Introduction

Potential Energy of a spring          PE  =  kx2  

Kinetic Energy                               KE  =  mv2  

Mechanical Energy of a spring        E  =  mv2   +   kx2  

Step (a)  at the amplitude, maximum displacement, the mass is motionless.

E = PE only

     

    

   

Step (b) at the equilibrium position has no PE since x = 0.  

E = KE only

   

   

    

Step (c) at the amplitude for the compression step has a velocity of 0.

E  =  PE only

   

     

   

    

Step (d) is between the equilibrium position and the amplitude.

E  =  PE  +  KE

   

    

   

      

The spring to the left stretches 0.150 m when a 0.300 kg mass is attached.  It stretches an additional 0.100 m from this equilibrium point and released.  Calculate (a) the spring constant, (b) the amplitude of the oscillation, (c) the maximum velocity, (d) the velocity when the mass is 0.050 m from equilibrium, and (e) the maximum acceleration of the mass.

(a) F = - kx       Solve for k  

        - F        - 9.80 m s -2  x  0.300 kg
k  =  ----  =  --------------------------------
          x                   - 0.150 m

k  =  19.6 N/m

(b)  A  =  0.100 m       The amount the spring was stretched from equilibrium position after mass was added.  

(c)  KE max  =  PE max        mv2  =  kA2         Solve for v.  
   

                  k A2                    19.6 N/m x ( 0.100 m )2 
      v  =  ( --------- ) 1/2  =  ( -------------------------------- ) 1/2  =  0.808 m/s
                     m                                   0.300 kg
    

(d)  E =   mv2   +   kx2   =  kA2        Solve for v.  

       v  =  ( k ( A2  -  x2 ) / m ) 1/2  =  ( 19.6 N/m ( (0.100 m)2  -  (0.050 m)2 ) / 0.300 kg ) 1/2  =  0.700 m/s

        
(e)  F  =  ma  =  kA        Greatest force is at the amplitude.  Solve for acceleration.   

      
                 kA         19.6 N/m x 0.100 m
       a  =  -------  =  ----------------------------  =  6.53 m/s2     
                   m                    0.300 kg
    

Table of Contents

     

Problems

1.     A 750. g mass at rest on the end of a horizontal spring ( k = 124 N/m ) is struck by a hammer which gives the mass an initial velocity of 2.76 m/s.  Calculate (a) the amplitude, (b) the maximum acceleration, and (c) the total energy.
    
2. A mass sitting on a horizontal, frictionless surface is attached at one end to a spring; the other end is fixed to a wall.   3.00 J of work are required to compress the spring by 0.125 m.  If the mass is released from rest with the spring compressed, it experiences a maximum acceleration of 15.5 m s -2.  Find the value of (a) the spring constant, and (b) the mass.
   
3. (a) At what displacement from equilibrium is the energy of a SHO half KE and half PE?  (b) At what displacement from equilibrium is the speed half the maximum value?
   
4. A mass on the end of a spring is stretched a distance xo from equilibrium and released.  At what distance from equilibrium will it have an acceleration equal to half its maximum acceleration?

Table of Contents

    

Answers

1.     A 750. g mass at rest on the end of a horizontal spring ( k = 124 N/m ) is struck by a hammer which gives the mass an initial velocity of 2.76 m/s.  Calculate (a) the amplitude, (b) the maximum acceleration, and (c) the total energy.
    
   
mv 2  =  kA2              At initial velocity, there is no potential energy.  At Amplitude no KE.  Solve for A.

   
            mv2                 0.750 kg x ( 2.76 m s -1 )2 
A  =  ( ------- ) 1/2   =  ----------------------------------  =  0.214 m  (a)
              k                            124 N/m
    

(b) The maximum acceleration occurs with the maximum force at the amplitude.
   
F  =  ma  =  - kA           Solve for a.
   

         - k A          - 124 N/m x 0.214 m
a  =  ---------  =  ----------------------------  =  - 35.4 m s -2     Acceleration is in opposite direction of displacement.
            m                       0.750 kg
   

(c)  At the amplitude, the total energy = potential energy.

E  =  kA2  =  x 124 N/m x ( 0.214 m ) 2  =  2.84 J
    
   

   
2. A mass sitting on a horizontal, frictionless surface is attached at one end to a spring; the other end is fixed to a wall.   3.00 J of work are required to compress the spring by 0.125 m.  If the mass is released from rest with the spring compressed, it experiences a maximum acceleration of 15.5 m s -2.  Find the value of (a) the spring constant, and (b) the mass.
   
   
(a)  Work = kA2             Solve for k.

     
          2 x Work          2 x 3.00 J
k  =  ---------------  =  -------------------  =  384 N/m
                A 2              ( 0.125 m )2  
   

(b)  F  =  ma  =  - kA                 Maximum acceleration and force occur at the amplitude.  Solve for m.

    
          - kA         - 384 N/m x ( - 0.125 m) 
m  =  -------  =  ----------------------------------  =  3.10 kg
             a                     15.5 m s -2  
    
    
   

   
3. (a) At what displacement from equilibrium is the energy of a SHO half KE and half PE?  (b) At what displacement from equilibrium is the speed half the maximum value?
   
   
(a)  PE =  kA2           At the amplitude all of the energy is Potential Energy.

PE  =   kx2        x is the displacement when the energy is half potential energy and half kinetic energy.

0.25 kA2   =  kx2           Solve for x.
           
x  =  (  0.5 A2  ) 1/2  =  0.707 A
          

(b)    kA2  =   mv2          This is the maximum velocity.   Solve for velocity.

      
           kA2                                              kA2                 kx2 
v  =  ( ------ ) 1/2        thus      0.5v  =  ( ------- ) 1/2  =  ( ------ ) 1/2         Solve for displacement, x.
             m                                                4m                    m

   
                                      kA2                         kA2  
kA2  =  kx2   +  m -------  =   kx2   +  --------               E  =  PE  +  KE       E = PE at amplitude
                                      4m                           8 
    

kA2  =  kx2  +  1/4 kA2                                    Solve for x.

kx2  =  3/4  kA2   
    

             3A2 
x  =  ( -------- ) 1/2   =  0.866 A
                  4
    
   

   
4. A mass on the end of a spring is stretched a distance xo from equilibrium and released.  At what distance from equilibrium will it have an acceleration equal to half its maximum acceleration?
    
   
F  =  ma  =  - kA         At the maximum force which is also the maximum accelertion.     
     

            - kA
ma  = --------  =  -kx            One-half the maximum acceleration is at one-half the force.  Solve for x.
              2 
   

          A
x  =  -----  =  0.5 A
          2
   

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