2 Couple Pndulum

8/12/2019 2 Couple Pndulum

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Vibrations and Waves MP205, Assignment 6 Solutions

* 1. Two identical pendulums are connected by a light coupling spring. Eachpendulum has a length of 0.4 m, and they are at a place where g =9.8 ms2. With the coupling spring connected, one pendulum is clamped

and the period of the other is found to be 1.25 sec exactly.

(a) With neither pendulum clamped, what are the periods of the twonormal modes?

(b) What is the time interval between successive maximum possible am-plitudes of one pendulum after one pendulum is drawn aside andreleased?

(Ill go through this first question in quite some detail)

(a) Our system is given by:

We can write a general equation for A, by looking at the case where we moveA a distance xA, and B a distance xB and seeing what forces affect A as aresult.

There are two forces affecting Ahere:

restoring force due to the pendulum

a=

20x

ma= m20xF = m2

0x

restoring force due to the spring

F = kx

where x is the change in spring length: xA xB


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So the total force on Ais:

FA = m20xA k(xA xB)

md2xAdt2

= m20xA k(xA xB)d2xAdt2 =

2

0xA k

m (xA xB)setting c=

km

we get:

d2xAdt2

+20

xA+2c (xA xB) = 0

Similarly, the equation of motion for B is:

d2xBdt2

+20

xB+2

c (xB xA) = 0

Were told that whenBis clamped (xB = 0), that the period ofAisTA = 1.25s.

TA= 2A

A= 2

TA=

2

1.25= 5.03

2A= (5.03)2 = 25.27

Looking at this system:

d2xAdt2

+20

xA+2c (xA xB) = 0

d2xAdt2

+20xA+2

c (xA 0) = 0d2xA

dt2

+20xA+2cxA= 0

d2xAdt2

+ (20

+2c )xA= 0

A=

20+ 2c

For a simple pendulum of length l, the angular frequency0 is given by:

0=

g

l

20 =g

lSo in our case:

20 =9.8

0.4= 24.5


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Using this in our expression for A gives:

A=

20+ 2c

A=

24.5 +2c2A= 24.5 +

2c

25.27 = 24.5 +2c2c = 25.27 24.5 = 0.77c= 0.88

To find the normal modes we let xA = Ccos(t) and xB = D cos(t) andobtain an expression for that satisfies both equations of motion.:

xA=Ccos(t) d2xA

dt2 = 2Ccos(t)

xB =D cos(t)

d2xB

dt2 = 2

D cos(t)

Using this in our equations of motion:

First for A:d2xAdt2

+20xA+2

c (xA xB) = 02Ccos(t) +2

0Ccos(t) +2c (Ccos(t) D cos(t)) = 0

2C+20C+2c (C D) = 0C(2 +20+ 2c ) 2cD= 0

CD

= 2

c

2 +20+ 2cAnd for B:

d2xBdt2

+20xB+ 2

c (xB xA) = 02D cos(t) +20D cos(t) +2c (D cos(t) Ccos(t)) = 0

2D+20D+2c (D C) = 0D(2 +20+ 2c ) 2cC= 0

C

D=2 +2

0+ 2c

2c


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Combining these results:

CD

= C

D2c

2 +20+ 2c=2 +2

0+2c

2c

(2

c )2

= (2

+2

0+ 2

c )2

2c = 2 +20+2c2 =2

0+2c 2c

2 =20

+2c +2c

=20

+ 22c

=

20+ 22c

=

24.5 + 2(0.77) =

24.5 + 2(0.77) =

26.04 = 5.1

T =2

=

2

5.1

= 1.23s

2 =20+2c 2c=2

0

=0=

24.5 = 4.95

T =2

=

2

4.95= 1.27s

Note: normal modes for this system are given by the following two cases:

(i) The case whereA and B are pulled apart the same distance: xA= xB.

(ii) The case whereAandB are pulled in the same direction the same distance:xA = xB.

Looking at (i), and the equation of motion for A:

d2xAdt2

+20

xA+2c (xA xB) = 0

d2xAdt2

+20xA+2

c (xA+xA) = 0

d2xAdt2

+20

xA+ 22cxA= 0

d2

xAdt2

+ (20+ 22c )xA= 0

=

20+ 22c


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Looking at (ii), and the equation of motion for A:

d2xAdt2

+20

xA+2

c (xA xB) = 0d2xAdt2

+20xA+2c (xA xA) = 0

d2xAdt2

+20

xA = 0

= 0

(b) Starting with our general equations of motion:

d2xAdt2

+20xA+2c (xA xB) = 0

d2xBdt2

+20

xB+2c (xB xA) = 0

Adding these equations:

d2(xA+xB)

dt2 +2

0(xA+xB) = 0

and setting q1= xA+xB we get:

d2q1dt2

+20

q1= 0

so we can write q1 as:

q1=Ccos(0t) =Ccos(4.95t)

Subtracting these equations:

d2(xA xB)dt2

+20(xA xB) + 22c (xA xB) = 0d2(xA xB)

dt2 + (2

0+ 22c )(xA xB) = 0

using =

20

+ 22c and q2= xA xB we get:d2q2dt2

+2q2= 0

so we can write q2 as:

q2=D cos(t) =D cos(5.1t)

We can write xA and xB in terms ofq1 and q2.

q1=xA+xB q1= xA+xB

q2=xA xB q2= xA+xB

q1+q2= 2xA q1 q2= 2xBxA=

1

2(q1+q2) xB =

1

2(q1 q2)


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This gives us:

xA(t) =1

2(q1+q2)

=1

2

(Ccos(4.95t) +D cos(5.1t))

xB(t) =1

2(q1 q2)

=1

2(Ccos(4.95t) D cos(5.1t))

These are the general solutions for xA and xB. To go any further we needinitial conditions.In this case, were told one pendulum is drawn aside and released. This cor-responds to one pendulum having displacement 0 at t = 0, and one pendulumhaving some displacement (say A0) at t= 0.

Lets assume A is drawn aside, so at t = 0:

xA(0) =A01

2(Ccos(0) +D cos(0)) =A0

C+D = 2A0

xB(0) = 0

1

2(Ccos(0) D cos(0)) = 0

CD= 0C=D

Filling this into xA(0) we get:

C+C= 2A0

C=A0

D= C= A0.

This gives us our equations of motion:

xA(t) =1

2(A0cos(4.95t) A0cos(5.1t))

= A0

2 (cos(4.95t) cos(5.1t))

We can use the trigonometric identity cos A cos B = 2sin A+B2

sinAB2

= A0sin

(4.95 + 5.1)

2 t

sin

5.1 4.95

2 t

= A0sin (5t)sin(0.08t)

xB(t) =1

2(A0cos(4.95t) +A0cos(5.1t))

=A0

2

(cos(4.95t) + cos(5.1t))


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We can use the trigonometric identity cos A+ cos B = 2 cosA+B2

cosAB2

=A0cos

(4.95 + 5.1)

2 t

cos

5.1 4.95

2 t

=A0cos (5t)cos(0.08t)

xA(t) = A0sin (5t)sin(0.08t) xB(t) =A0cos (5t)cos(0.08t)

Were asked to find the time interval between maximum possible amplitudesof one pendulum in this system. Looking at xB we can note that this is simplya beat equation.If we plot xB against time we can see its motion:

Where the distance between peaks is the time interval between successive max-imum possible amplitudes.If we compare this to the motion of cos

0+

2 t

= cos(5t)

and the motion of cos0

2 t

= cos(0.08t)

We can see that the time interval between successive peaks is half the periodof the slow oscillating motion, which is given by:

T = 2|0|

2

= 4

|0 |


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So the time we require is the beat period:

T=1

2T =

2

|0 | = 2

|2(0.08)| = 39.3s

2. Two harmonic oscillators A and B, of mass m and spring constants kAand kB, respectively, are coupled together by a spring of spring constantkC. Show that the normal frequencies satisfy:

m2 =kA+kB

2 +kC

kA kB

2

2+k2C

Our equations of motion are:

d2xAdt2

+kAm

xA+kCm

(xA xB) = 0d2xB

dt2 +

kBm

xB+kCm

(xB xA) = 0

We want to find such that xA = Ccos(t) and xB = D cos(t) are solutions tothese equations.

2Ccos(t) + kAm

Ccos(t) +kCm

(Ccos(t) D cos(t)) = 0m2C+ kAC+ kC(C D) = 0

D

C =m2 +kA+kC

kC

2D cos(t) + kBm

D cos(t) +kCm

(D cos(t) Ccos(t)) = 0

m2

D+kBD+kC(D C) = 0D

C =

kCm2 +kB+kC

Combining these:

m2 +kA+kCkC

= kC

m2 +kB+kC(m2 +kA+kC)(m2 +kB+kC) =k2C

(m2)2 +kAkB+kAkC+ kBkC+ k2Cm2kA m2kB 2m2kC=k2C

(m

2

)

2

m2

(kA+kB+ 2kC) + (kAkB+ kAkC+ kBkC) = 0


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This is just a quadratic equation for m2, so we can solve it using the quadraticformula:

m2 =((kA+kB+ 2kC))

(kA+kB+ 2kC)2 4(1)(kAkB+kAkC+kBkC)

2(1)

=

kA+kB+ 2kC

k2A+k

2B+ 4k

2C+ 2kAkB+ 4kAkC+ 4kBkC

4kAkB

4kAkC

4kBkC

2

=kA+kB

2 +kC

k2A+k

2B 2kAkB+ 4k2C

2

=kA+kB

2 +kC

(kA kB)2 + 4k2C

4

=kA+kB

2 +kC

kA kB4

2

2+k2C

As required.

3. Two equal masses on an effectively frictionless horizontal air track areheld between rigid supports by three identical springs, as shown. Thedisplacements from equilibrium along the line of the springs are describedby coordinates xA and xB, as shown. If either of the masses is clamped,the period Tfor one complete vibration of the other is 3 sec.

*(a) If both masses are free, what are the periods of the two normalmodes of the system?

(a) Our equations of motion are:

d2

xAdt2

+ km

xA+ km

(xA xB) = 0d2xAdt2

+ k

m(2xA xB) = 0

d2xBdt2

+ k

mxB+

k

m(xB xA) = 0

d2xBdt2

+ k

m(2xB xA) = 0

Setting 0=

km

:

d2xAdt2 +

20(2xA xB) = 0d2xB

dt2 +2

0(2xB xA) = 0


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Were told that when one mass is clamped, the period of the other is T= 3s.Lets clampB , soxB = 0. Using this in the equation of motion for A we obtain:

d2xAdt2

+20(2xA xB) = 0d2xA

dt2

+ 220

xA= 0

= 20We know the period of this system is T= 3s.

T =2

= 3

=2

320=

2

3

0=

2

3To obtain the normal modes we set xA = Ccos(t) and xB = D cos(t)andobtain an expression for that satisfies both equations of motion.

d2xAdt2

+20(2xA xB) = 02Ccos(t) +20(2Ccos(t) D cos(t)) = 0

2C+ 20(2C D) = 0C

D=

20220

2

d2xBdt2

+20

(2xB xA) = 02D cos(t) +2

0(2D cos(t) Ccos(t)) = 0

2D+20(2D C) = 0C

D=

220 2

20

Combining these:

20

220

2 =

220 2

20(20)

2 = (220 2)220220 2

2 = 220 20=0=

2

3

T=2

=

223

= 3

2s

=

30=

3

2

3 =

2

3T =

2

=

223

=

6s


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Note: It is easy to check that these normal modes coincide with the two fol-lowing cases:

=0Both masses pulled a distance xin the same direction.

= 30Both masses pulled a distance xin opposite directions.

At t = 0, mass A is at its normal resting position and mass B is pulledaside a distance of 5 cm. The masses are released from rest at this instant.

(b) Write an equation for the subsequent displacement of each mass asa function of time.

(c) What length of time (in seconds) characterises the periodic transferof the motion from B to A and back again?

(b) Starting with our equations of motion:

d2xAdt2

+20(2xA xB) = 0d2xBdt2

+20

(2xB xA) = 0Adding our equations of motion gives:

d2q1dt2

+20

q1= 0

where q1=xA+xB.

Subtracting our equations of motion gives:

d2q2dt2 + 3

2

0q2= 0where q2=xA xB.

SO we can write q1= Ccos(0t) andq2= D cos(

30t). As in question 1., wecan thus write xA and xB:

xA(t) =1

2

Ccos(0t) +D cos(

30t)

xB(t) =

1

2Ccos(0t) D cos(

30t)


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(b) Putting0 =

k0/m, show that the angular frequencies 1 and2 ofthe normal modes are given by

1 = 0 , 2 = [2

0+ (2kc/m)]

1/2,

and that the angular frequency of A when B is clamped (xB = 0

always) is given by

A = [2

0+ (kc/m)]1/2.

(c) Using the numerical data above, calculate the expected frequency(2) of the higher normal mode. (The observed value was2.27 sec

1).

(d) From the same data calculate the ratiokc/k0 of the two spring con-stants.

(a) Looking at our system:

Looking at A: There are two forces affecting A here:

restoring force due to the spring on its left

F = k0xA

where xA is the change in spring length, and k0 is the spring constant. restoring force due to the spring on its right

F = kC(xA xB)

wherexAxBis the change in spring length, andkCis the spring constant.So, the total force on A is (denoting the total force on AbyFA):

FA= k0xA kC(xA xB)using F =ma

maA= k0xA kC(xA xB)and a= d

2xdt2

md2xAdt2

= k0xA kC(xA xB)

As this system is totally symmetric in A and B we can write the equation ofmotion for B by interchanging Aand B inAs equation of motion:

md2xB

dt2 = k0xB kC(xB xA)


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(b) Setting0=

k0m

(after dividing across bym):

d2xAdt2

= 20xA kCm

(xA xB)d2xB

dt2 = 2

0xB kC

m(xB xA)

To obtain the normal modes we set xA = Ccos(t) and xB = D cos(t) andobtain an expression for that satisfies both equations of motion.

2Ccos(t) = 20Ccos(t) kCm

(Ccos(t) D cos(t))

2C= 20

C kCm

(C D)C

D =

kCm

20+ kCm 2

2D cos(t) =

20

D cos(t)

kC

m

(D cos(t)

Ccos(t))

2D= 20D kCm

(D C)C

D =

20

+ kCm 2

kCm

Combining these:kCm

20+ kCm 2 =

2 + kCm 20

kCm

(kC

m

)2 = (20

+kC

m2)2

kCm

=20+kCm 2

2 =20+

kCm kC

m 1= 0

2=

20+2kC

m

Note: It is easy to check that these normal modes coincide with the two fol-lowing cases:

=0Both masses pulled a distance xin the same direction.

=

20+ 2kCm

Both masses pulled a distance xin opposite directions.


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When B is clamped, xB = 0, using this in the equation of motion for A:

d2xAdt2

= 20

xA kCm

(xA xB)d2xAdt2

= 20xA kCm

xA

d2xAdt2

= (20+ k

C

m)xA

A =

20

+kCm

(c) Were told that A= 1.81s1 and 1= 1.14s1:

= 1

T =

2 = 2

1= 21= 2(1.14) = 7.161=0= 7.16

A= 2A= 2(1.81) = 11.37

A=

20+

kCm

= 11.37

20

+kCm

= 129.34

(7.16)2 +kCm

= 129.34

k

m = 129.34 51.27 = 78.07Now obtaining a numerical value for 2 is simple:

2=

20+

2kCm

=

(7.16)2 + 2(78.07) =

207.41 = 14.4

Using this in = 2

gives us:

2= 22

=14.4

2 = 2.29s1

(d) We want the ratio kCk0

:

kCk0

=kCmk0m

=kCm

20=

78.07

51.27= 1.52

2 Couple Pndulum

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