Homework

1. Tutorial-3
Coupled Oscillator
Problem 1
There are two small massive beeds, each of mass 𝑀, on a taut massless string of
length 8𝐿, as shown. The tension in the string is 𝑇.
1. Derive expressions for the normal mode angular frequencies of the system,
for small displacements of the system in a plane. Assume that 𝑇 is
suffieciently large that you may ignore gravity.
2. If at 𝑡 = 0 both masses are stationary, one is at equilibrium and the other
displaced from equilibrium by a distance 𝐴, write an expression for
displacement 𝑦1(𝑡) as a function of time of the mass initially in the
equilibrium position.
Solution:
1) Masses cannot be displaced in horizontal direction (x-direction). They can
be moved in vertical direction only.
Tension = 𝑇,

2. The force equations are as follows.
Particle (bead) 1:
𝑚
𝑑2𝑦1
𝑑𝑡2
= −
𝑇
3𝐿
𝑦1 −
𝑇
2𝐿
(𝑦1 − 𝑦2) (1)
Particle (bead) 2:
𝑚
𝑑2𝑦2
𝑑𝑡2
= −
𝑇
3𝐿
𝑦2 −
𝑇
2𝐿
(𝑦2 − 𝑦1) (2)
Normal modes can be found by writing 𝑦1 and 𝑦2 in linear combination
form.
𝑞1 =
𝑦1+𝑦2
2
and 𝑞2 =
𝑦1−𝑦2
2
After subtracting and adding the values of 𝑞1 and 𝑞2we can rewrite equation
(1) and (2) as follows written as:
𝑚
𝑑2𝑞1
𝑑𝑡2
= −𝑘𝑞1 (3)
and 𝑚
𝑑2𝑞2
𝑑𝑡2
= −(𝑘+2𝑘′
)𝑞2 (4)
where, 𝑘 =
𝑇
3𝐿
and 𝑘′
=
𝑇
2𝐿
∴ 𝜔1 = √
𝑘
𝑚
= √
𝑇
3𝐿𝑚
and 𝜔2 = √
𝑘+2𝑘′
𝑚
= √
4𝑇
3𝐿𝑚
So, the expressions for the normal mode angular frequencies are 𝜔1 = √
𝑇
3𝐿𝑚
and 𝜔2 = √
4𝑇
3𝐿𝑚
.
2) Solutions for equation (3) and (4) are
𝑞1 =
𝑦1+𝑦2
2
= 𝐴1𝑐𝑜𝑠(𝜔1𝑡) (5)
𝑞2 =
𝑦1−𝑦2
2
= 𝐴2𝑐𝑜𝑠(𝜔2𝑡) (6)
From equation (5),

3. 𝑦1 + 𝑦2
2
= 𝐴1𝑐𝑜𝑠(𝜔1𝑡)
at 𝑡 = 0, 𝑦1 = 𝐴 and 𝑦2 = 0
∴ 𝐴1 = 𝐴/2
Similarly, 𝐴2 = 𝐴/2
So,
𝑞1 = 𝐴/2𝑐𝑜𝑠(𝜔1𝑡) (7)
𝑞2 = 𝐴/2𝑐𝑜𝑠(𝜔2𝑡) (8)
Solving equation (5) and (6), we get
𝑦1(𝑡) =
𝐴
2
[𝑐𝑜𝑠(𝜔1𝑡) + 𝑐𝑜𝑠(𝜔2𝑡)]
where, 𝜔1 = √
𝑇
3𝐿𝑚
and 𝜔2 = √
4𝑇
3𝐿𝑚
.
Problem 2
The figure above shows a system of masses in the x-y plane. The mass of 2𝑚 is
connected to an immobile wall with a spring of constant 2𝑘, while the mass of 𝑚 is
connected to an immobile wall with a spring of constant 𝑘. The masses are then
coupled to each other with an elastic band of length 𝐿, under tension 𝑇 = 2𝑘𝐿. The
masses are constrained to move in the x direction only. At equilibrium the masses
have the same x position and the springs are uncompressed. There is no friction or

4. gravity. The displacements from equilibrium are small enough (𝑥1, 𝑥2 ≪ 𝐿), so that
the tension in the band stays constant.
1. Find the normal mode frequencies of the system.
2. What are the ratios of the displacements from equilibrium for two different
normal modes? (Hint: Basically, you have to find out the eigen functions for
corresponding normal mode frequencies)
Solution:
1) Equation of motions for 2𝑚 and 𝑚 :
2𝑚
𝑑2𝑥1
𝑑𝑡2
= −2𝑘𝑥1 −
𝑇
𝐿
(𝑥1 − 𝑥2) (1)
and 𝑚
𝑑2𝑥2
𝑑𝑡2
= −𝑘𝑥2 −
𝑇
𝐿
(𝑥2 − 𝑥1) (2)
where, 𝑇 = 2𝑘𝐿 and 𝐿 is length of band.
From equation (1) and (2),
𝑚
𝑑2𝑥1
𝑑𝑡2
+ 2𝑘𝑥1 − 𝑘𝑥2 = 0 (3)
and 𝑚
𝑑2𝑥2
𝑑𝑡2
− 2𝑘𝑥1 + 3𝑘𝑥2 = 0 (4)
Say this coupled oscillator oscillating with a coupled frequency 𝜔 then
solution to these equations are
𝑥1 = 𝐴𝑒𝑖𝜔𝑡
and 𝑥2 = 𝐵𝑒𝑖𝜔𝑡
Substituting these above two expressions in (3) and (4) the following matrix
equation :
[−𝑚𝜔2
+ 2𝑘 −𝑘
−2𝑘 −𝑚𝜔2
+ 3𝑘
] [
𝐴
𝐵
] = [
0
0
] (5)
𝜔 can be found by solving eigenvalue equation,

5. (−𝑚𝜔2
+ 2𝑘)(−𝑚𝜔2
+ 3𝑘) − 2𝑘2
= 0
or, 𝑚2
𝜔4
− 5𝑘𝑚𝜔2
+ 4𝑘2
= 0
or, 𝜔2
=
5𝑘𝑚±√25𝑘2𝑚2−16𝑘2𝑚2
2𝑚2
or, 𝜔2
=
5𝑘±3𝑘
2𝑚
=
𝑘
𝑚
,
4𝑘
𝑚
∴ 𝜔1 = √
𝑘
𝑚
and 𝜔2 = √
4𝑘
𝑚
.
2) Eigen functions can be calculated usin (5) for 𝝎𝟏 = √
𝒌
𝒎
[
𝑘 −𝑘
−2𝑘 2𝑘
] [
𝐴
𝐵
] = [
0
0
]
or, 𝐴 − 𝐵 = 0; or, 𝐴 = 𝐵
∴ we can write.
𝒙𝟏
𝒙𝟐
= 𝟏
So, 𝒙𝟏 = 𝟏 and 𝒙𝟐 = 𝟏; (say)
Again, for 𝝎𝟐 = √
𝟒𝒌
𝒎
[
2𝑘 −𝑘
−2𝑘 −𝑘
] [
𝐴
𝐵
] = [
0
0
]
or, −2𝑘𝐴 − 𝐵 = 0; or, 𝐵 = −2𝐴
∴
𝒙𝟐
𝒙𝟏
= −𝟐
So, 𝒙𝟏 = 𝟏 and 𝒙𝟐 = −𝟐; (say)
Problem 3
Two identical spring-mass systems 𝐴 and 𝐵 are coupled by a weak middle spring
having a spring constant smaller by a factor of 100 (i.e. 100𝑘𝑚𝑖𝑑𝑑𝑙𝑒 = 𝑘𝐴 = 𝑘𝐵).

6. Mass 𝐴 is pulled by a small distance and released from rest, while mass 𝐵 is released
from rest at its equilibrium position, at 𝑡 = 0. Calculate the approximate number of
oscillations completed by mass 𝐴 before its oscillations die down first.
Solution:
Force equations for both the masses:
𝑚1
𝑑2𝑥1
𝑑𝑡2
= 𝐹1 = −𝑘1𝑥1 − 𝑘′
(𝑥1 − 𝑥2) (1)
and
𝑚2
𝑑2𝑥2
𝑑𝑡2
= 𝐹2 = −𝑘2𝑥2 − 𝑘′
(𝑥2 − 𝑥1) (2)
From equation (1) and (2), we can write in matrix form,
(
𝑚1 0
0 𝑚2
)
(
𝑑2
𝑥1
𝑑𝑡2
𝑑2
𝑥2
𝑑𝑡2 )
= − (
𝑘1 + 𝑘′
−𝑘′
−𝑘′
𝑘2 + 𝑘′) (
𝑥1
𝑥2
)
or, 𝑀
̂𝑋̈
̂ = −𝐾
̂𝑋
̂
or, 𝑋̈
̂ = −𝑀
̂−1
𝐾
̂𝑋
̂ (3)
Where, 𝑀
̂, 𝑋̈
̂, 𝐾
̂ and 𝑋
̂ all are matrix.
Now, Let us consider a normal mode solution,

7. 𝑋
̂ = 𝐴
̂𝑒𝑖𝜔𝑡
(4)
or,
𝑋̈
̂ = −𝜔2
𝑋
̂ (5)
Putting the value of 𝑋̈
̂ from equation (3)
or, 𝑀
̂−1
𝐾
̂𝑋
̂ = 𝜔2
𝑋
̂
or, 𝑀
̂𝑀
̂−1
𝐾
̂𝑋
̂ = 𝑀
̂𝜔2
𝑋
̂
or, 𝐾
̂𝑋
̂ = 𝑀
̂𝜔2
𝑋
̂ (6)
Calculated above, 𝑀
̂ = (
𝑚 0
0 𝑚
); 𝑚1 = 𝑚2 = 𝑚
𝐾
̂ = (
𝑘 + 𝜀𝑘 −𝜀𝑘
−𝜀𝑘 𝑘 + 𝜀𝑘
);
where, 𝑘1 = 𝑘2 = 𝑘 & 𝑘′
= 𝜀𝑘 = 𝑘/100
After solving (6) and normalized, we have
𝜔1 = √
𝑘
𝑚
; 𝐴1 =
1
√2
(
1
1
)
and 𝜔2 = √(1 + 2𝜀)
𝑘
𝑚
; 𝐴2 =
1
√2
(
1
−1
)
The most most general solution of coupled oscillator is,
𝑋
̂(𝑡) = 𝐶1𝐴
̂1 cos(𝜔1𝑡) + 𝐶2𝐴
̂2 cos(𝜔2𝑡) (7)
At 𝑡 = 0, mass A is released from a small distance = 𝑑 and mass B kept at
rest. So,
𝑋
̂(0) = (
𝑥1(0)
𝑥2(0)
) = (
𝑑
0
)
From equation (7),

8. (
𝑑
0
) =
𝐶1
√2
(
1
1
) +
𝐶2
√2
(
1
−1
),
or, 𝐶1 = 𝐶2 = 𝑑/√2
So, equation (7) becomes,
𝑋
̂(𝑡) = 𝐶1𝐴
̂1 cos(𝜔1𝑡) + 𝐶2𝐴
̂2 cos(𝜔2𝑡)
or, (
𝑥1(𝑡)
𝑥2(𝑡)
) =
𝐶1
√2
(
1
1
) cos(𝜔1𝑡) +
𝐶2
√2
(
1
−1
) cos(𝜔2𝑡)
or, (
𝑥1(𝑡)
𝑥2(𝑡)
) =
𝑑
2
(
1
1
) cos(𝜔1𝑡) +
𝑑
2
(
1
−1
) cos(𝜔2𝑡)
Comparing both side,
∴ 𝑥1(𝑡) =
𝑑
2
[cos(𝜔1𝑡) + cos(𝜔2𝑡)] = 𝑑 𝑐𝑜𝑠 (
𝜔2 − 𝜔1
2
𝑡) 𝑐𝑜𝑠 (
𝜔1 + 𝜔2
2
𝑡)
And
𝑇 = Time period of the envelop =
2𝜋
𝜔2−𝜔1
2
=
4𝜋
𝜔2−𝜔1
∴ Number of oscillation in time
𝑇
4
=
𝑇
4
×
𝜔1+𝜔2
2×2𝜋
=
1
4
×
4𝜋
𝜔2−𝜔1
×
𝜔1+𝜔2
4𝜋
=
1
4
(
𝜔1 + 𝜔2
𝜔2 − 𝜔1
)

9. =
1
4
×
1 + √1 + 2𝜀
√1 + 2𝜀 − 1
=
1
4
×
√1 + 2/100 + 1
√1 + 2/100 − 1
=
1
4
×
√1 + 1/50 + 1
√1 + 1/50 − 1
= 50.49 ≈ 50
Problem 4
Three identical masses 𝑚 are connected in series to four identical springs each of
force constant 𝑘 (each mass coupled to two strings, and the end springs to walls):
a) Calculate the eigenfrequencies of small oscillations (normal modes) of this
coupled oscillator system.
b) What sets of initial displacements would you give to the three masses so that
the system oscillates in each of these normal modes.
Solution:
a) Equation of motions:
Mass 1→ 𝑚
𝑑2𝑥1
𝑑𝑡2
= −𝑘𝑥1 − 𝑘(𝑥1 − 𝑥2) (1)

10. Mass 2→ 𝑚
𝑑2𝑥2
𝑑𝑡2
= −𝑘((𝑥2 − 𝑥1) − 𝑘(𝑥2 − 𝑥3) (2)
Mass 3→ 𝑚
𝑑2𝑥3
𝑑𝑡2
= −𝑘𝑥3 − 𝑘(𝑥3 − 𝑥2) (3)
Say, all these masses oscillating with uniform frequency 𝜔; then
𝑥1 = 𝐴𝑒𝑖𝜔𝑡
; 𝑥2 = 𝐵𝑒𝑖𝜔𝑡
and 𝑥3 = 𝐶𝑒𝑖𝜔𝑡
(4)
Substituting these we get the following matrix equation;
[
−𝑚𝜔2
+ 2𝑘 −𝑘 0
−𝑘 −𝑚𝜔2
+ 2𝑘 −𝑘
0 −𝑘 −𝑚𝜔2
+ 2𝑘
] [
𝐴
𝐵
𝐶
] = [
0
0
0
]
Solving for 𝜔,
(−𝑚𝜔2
+ 2𝑘)[(−𝑚𝜔2
+ 2𝑘)2
− 𝑘2] + 𝑘[−𝑘(−𝑚𝜔2
+ 2𝑘)] = 0
or, (−𝑚𝜔2
+ 2𝑘)[(−𝑚𝜔2
+ 2𝑘)2
− 2𝑘2] = 0
or, (−𝑚𝜔2
+ 2𝑘) = 0, or, 𝜔2
= 2𝑘/𝑚, or, 𝝎𝟏 = √𝟐𝒌/𝒎 (4)
and (−𝑚𝜔2
+ 2𝑘)2
− 2𝑘2
= 0
or, 𝑚2
𝜔4
− 4𝑚𝑘𝜔2
+ 2𝑘2
= 0
or, 𝜔2
=
4𝑚𝑘±√16𝑚2𝑘2−8𝑚2𝑘2
2𝑚2
or, 𝝎𝟐,𝟑 = √(𝟐±√𝟐)𝒌
𝒎
(5)
b) Eigen states: For 𝝎𝟏 = √𝟐𝒌/𝒎,
[
0 −𝑘 0
−𝑘 0 −𝑘
0 −𝑘 0
] [
𝐴
𝐵
𝐶
] = [
0
0
0
]
or, 𝐵 = 0 and 𝐴 = −𝐶 = 1 (say)

11. at 𝑡 = 0;
𝑥1 = −𝑥3 and 𝑥2 = 0
So, 𝑥1(0) = 1, 𝑥2(0) = 0 and 𝑥3(0) = −1
For 𝝎𝟐 = √(𝟐 + √𝟐)𝒌/𝒎,
[
√2𝑘 −𝑘 0
−𝑘 √2𝑘 −𝑘
0 −𝑘 √2𝑘
] [
𝐴
𝐵
𝐶
] = [
0
0
0
]
or, √2𝑘𝐴 − 𝐵𝑘 = 0 or, √2𝐴 = 𝐵;
−𝐵𝑘 + √2𝑘𝐶 = 0 or, √2𝐶 = 𝐵
and −𝐴 + √2𝐵 − 𝐶 = 0
So, 𝐴 = 𝐶 and 𝐴 = 𝐶 = 𝐵/√2
Let, 𝐴 = 𝐶 = 1, so, 𝐵 = √2
So, at 𝑡 = 0;
𝑥1 = 𝑥3 and 𝑥2 = 𝐵𝑒𝑖𝜔𝑡
So, 𝒙𝟏(𝟎) = 𝒙𝟑(𝟎) = 𝟏 and 𝒙𝟐(𝟎) = √𝟐
Similarly, for 𝝎𝟑 = √(𝟐 − √𝟐)𝒌/𝒎,
So, 𝒙𝟏(𝟎) = 𝒙𝟑(𝟎) = 𝟏 and 𝒙𝟐(𝟎) = −√𝟐