PHY
1150
Chapter 13; Oscillations
about Equilibrium
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D13.1 Find the frequency of a mass on a spring oscillator with the following values for the mass and the spring constant:
mass (kg) 
spring constant ( N / m ) 

a 
1.0 
2.5 
b 
2.5 
1.0 
c 
3.0 
7.2 
d 
50 
100 
e 
80 
200 
f 
100 
250 
g 
120 
250 
h 
200 
500 
This is an excellent one to solve with a spreadsheet:
D13.2 Find the frequency of the oscillator in this arrangement with the mass attached between two springs of spring constants k _{1} and k _{2}.
The next arrangement of springs is a
little more "tricky" or more interesting. Find the frequency of the oscillator
in this arrangement with the mass attached to two springs of spring constants
k _{1}
and k _{2}
as shown in the diagram here.
And with this third arrangement, again, find the frequency of the oscillator with the mass attached to two springs of spring constants k _{1} and k _{2} as shown in the diagram here.
_{For any simple harmonic oscillator, we have}
_{}
where k in this equation is the "effective
spring constant", the spring constant of a single spring that has
the same effect as the combination of springs in the diagram.
With this arrangement, a displacement of x causes a force of F_{1} =  k _{1} x due to the spring on the left and a force of F_{2} =  k _{2} x due to the spring on the right.The net force, then, is F_{net} = F_{1} + F_{2} =  k _{1} x  k_{ 2} x =  (k _{1} + k _{2})x. Since the effective spring constant, k_{eff} is the constant is F_{net} =  k _{eff }x, we can see that
k _{eff} = k _{1} + k_{ 2} The next arrangement of springs is a little more "tricky" or more interesting:
When the mass is moved a distance x, spring #1 stretches a distance x _{1} and spring #2 stretches a distance x _{2} with
x = x _{1} + x _{2} That is, the two springs need not be stretched (or compressed) the same amount at all. However, the force exerted by each spring must be the same. That is,
F_{1} = F_{2} or
k _{1} x _{1} = k _{2} x _{2} or
x_{1} = ( k _{2} / k _{1} ) x _{2} or
x _{2} = ( k _{1} / k _{2} ) x _{1} To determine the "effective spring constant", we must write the force in the form of
F =  k _{eff } x We have
F =  k _{1} x _{1} and
x = x _{1} + x_{ 2}
x = x _{1} + ( k _{1} / k _{2} ) x _{1}
x = [ 1 + ( k _{1} / k _{2} )] x _{1}
Therefore,
Therefore,
which can also be written as
The third arrangement of springs is actually fairly easy and straightforward. With this arrangement, a displacement of x causes a force of F_{1} =  k_{1}x due to spring #1 and a force of F_{2} =  k_{2}x due to spring #2. The stretch of each spring is the same as the displacement of the mass.
The net force, then, is F_{net} = F_{1} + F_{2} =  k _{1 }x  k _{2 }x =  ( k _{1} + k _{2 }) x. Since the effective spring constant, k _{eff} is the constant is F_{net} =  k _{eff }x , we can see that
k_{ eff} = k _{1} + k _{2}
D13.3 How much work is done when a spring with spring constant 80 N/m is stretched 0.15 m (or 15 cm)?W = (^{1}/_{2}) k x^{2}
W = (^{1}/_{2}) (80 N/m) (0.15 m)^{2}
W = 0.9 J
As the mass moves through equilibrium,
x = 0
so it has zero potential energy and its total energy is now all KE,
KE = (^{1}/_{2}) m v ^{2} = (^{1}/_{2}) (0.2 kg) v ^{2} = 0.25 J = E
v^{2} = 2.5 m ^{2} / s ^{2}
v = 1.58 m / s
D13.5 An object undergoes simple harmonic motion with an amplitude of 12 cm. At a point 8.0 cm from equilibrium, its speed is 20 cm/s. What is the period?
Caution: be very careful with the units!
We have used lower case m for both mass and meters. That gets confusing here so I will go to upper case M for mass.
E = PE _{max} = (^{1}/_{2}) k A^{2} A = 12 cm = 0.12 m
E = (^{1}/_{2}) ( k ) (0.12 m)^{2}
E = KE + PE = (^{1}/_{2}) M v^{2} + (^{1}/_{2}) k x^{2}
E = (^{1}/_{2}) M (0.2 m/s)^{2} + (^{1}/_{2}) k (0.08 m)^{2} = (^{1}/_{2}) ( k ) (0.12 m)^{2} = E
M (0.2 m/s)^{2} = k (0.12 m)^{2}  k (0.08 m)^{2}
(M/k)(0.2 m/s)^{2} = (0.12 m)^{2}  (0.08 m)^{2}
(M/k)(0.04) (m/s)^{2} = (0.0144  0.0064) m^{2} = 0.0080 m^{2}
(M/k) = 0.2 ( s^{2 })
T = 2 SQRT(M/k)
T = 2.8 sec
D13.6 For each of the simple harmonic oscillators listed in the table below, find the velocity of the mass at the position given:
These multipledataset problems are good ones to solve with a spreadsheet. This is basically energy conservation,
E = (^{1}/_{2}) k A^{2} = (^{1}/_{2}) k x^{2} + (^{1}/_{2}) m v^{2}
k A^{2} = k x^{2} + m v^{2}
m v^{2} = k (A^{2}  x^{2})
v^{2} = (m/k) (A^{2}  x^{2})
So this will be the formula we put in the spreadsheet:
f ^{2} = (^{ 1}/_{2} )
^{2} (k/m)
f ^{2} = ( ^{1}/_{4} ^{2})(k/m)
k = 4 ^{2}
f ^{2} m
k = 4 ^{2}
[2 (^{1}/_{s })] ^{2}
(0.050 kg)
k = 3.95 kg/s^{2}
k = 3.95 N/m
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(c) 2005, Doug Davis; all rights reserved