**PHY
1150**

__Chapter 15; Periodic
Motion__

**15.9, 11, 12, 15, 20, 25,
31, 34, 35**

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**15.9 This is an excellent one to solve with a
spreadsheet:**

**15.11 _{}**

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 thatk_{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}withx = x_{1}+ x_{2}

That is, the two springs neednotbe stretched (or compressed) the same amount at all. However, theforceexerted by each spring must be the same. That is,F_{1}= F_{2}

ork_{1}x_{1}= k_{2}x_{2}

orx_{1}= (k_{2}/k_{1}) x_{2}

orx_{2}= (k_{1}/k_{2}) x_{1}

To determine the "effective spring constant",we must write the force in the form ofF = - k_{eff}x

We haveF = - k_{1}x_{1}

andx = 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 thatk_{eff}= k_{1}+ k_{2}

15.12W = (^{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 haszeropotential energy and its total energy is now 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

**
**

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 (1/s^{2})

T = 2 «

T = 2.8 sec

**
**

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}**

f

k = 4

k = 4

k = 3.95 kg/s

k = 3.95 N/m

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(c) 2000, Doug Davis; all rights reserved