This document presents information about free damped harmonic motion. It defines simple harmonic motion and discusses damped versus free damped motion. Free damped motion has constant amplitude over time. The document provides an example differential equation to model free damped motion and solves it for a given initial value problem. The solution is found to be x(t) = 8cos(6t). The period and frequency of the resulting motion are calculated.

1. 4/4/2019 Department Of Mathematics
University Of Rajshahi1
Prepared By
MD. Bulbul Ahammed
ID:14044150
Group: B
Department Of Mathematics
University Of Rajshahi

2. PRESENTATION ON
FREE DAMPED HARMONIC MOTION BY MAKING A SUITABLE PROBLEM
Outlines
 Simple harmonic motion
 Damped and free damped motion
 Free damped motion with different cases.
 Free damped harmonic motion problem.
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3. Simple harmonic Motion
A system executing simple harmonic motion is called a harmonic oscillator. A harmonic
oscillator produces oscillations.
The oscillations can be of two types
i) Free damped harmonic motion/oscillations.
ii) Damped motion/oscillations.
Damped motion:
When the motion of an oscillator reduces due to an external force, the oscillator
and its motion are damped. As for example the motion is a simple pendulum.
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4. Free damped motion:
The oscillations whose amplitude remain constant with time are called free damped
motion/oscillation. As for example ,if the bob of a pendulum is displaced in vacuum and then
released, the bob will continue to execute Simple harmonic motion with constant amplitude.
Or, The sum of kinetic and potential energy remains a constant value.
 Damping Ratio: The damping ratio is a dimensionless measure describing how oscillations in a
system decay after a disturbance. It is denoted by a. If a=0, a<0, a=1, a>1 then the motion are
undamped, under damped, critically damped and over damped respectively.
Free damped motion with difference cases:
The differential equation for the motion of the mass on the spring is,
𝑚
𝑑2 𝑥
𝑑𝑡2 + 𝑎
𝑑𝑥
𝑑𝑡
+ 𝑘𝑥 = 𝐹(𝑡)
We now consider the special case of free damped motion, that is, in which both 𝑎 = 0 and 𝐹 𝑡 =
0 for all t.
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……………………(1)

5. Then from equ. (1), We have
𝑚
𝑑2
𝑥
𝑑𝑡2
+ 𝑘𝑥 = 0
Where 𝑚(> 0) is the mass and k(> 0) is the spring constant. Dividing through by 𝑚 and
letting
𝑘
𝑚
= 𝜔2
, We write (2) in the form
𝑑2
𝑥
𝑑𝑡2
+ 𝜔2 𝑥 = 0
The auxiliary equation
𝑟2
+ 𝜔2
= 0
has roots 𝑟 = ±𝜔𝑖 and hence the general solution of (2) can be written
𝑥 = 𝑐1 𝑠𝑖𝑛𝜔𝑡 + 𝑐2 𝑐𝑜𝑠𝜔𝑡
Where 𝑐1 and 𝑐2 are arbitrary constants.
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..…………..(2)
.………….…(3)
.……………(4)

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Free damped harmonic motion problem.
Solve and interpret the IVP. Also find the period and frequency
𝑑2 𝑥
𝑑𝑡2
+ 36𝑥 = 0, 𝑥 0 = 8, 𝑥′ 0 = 0;

7. Solution: Given that,
𝑑2
𝑥
𝑑𝑡2
+ 36𝑥 = 0,
Comparing this with 𝑎𝑟2
+ 𝑏𝑟 + 𝑐 = 0, We see that 𝑎 = 1, 𝑏 = 0, 𝑐 = 36.
Therefore, from (i)
𝑟2+36 = 0
𝑜𝑟, 𝑟2
= −36
∴ 𝑟 = 0 ± 6𝑖
This are 𝛼 ± 𝑖𝛽 form, So the general solution of (i) is,
𝑥 𝑡 = 𝑐1 cos 6𝑡 + 𝑐2 sin 6𝑡
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…………………..(i)
….……………..(ii)

8. Now imposing the conditions in (ii)
𝑥 0 = 𝑐1 cos 0 + 𝑐2 sin 0
⇒ 8 = 𝑐1 + 0
∴ 𝐶1 = 8
Putting the value of 𝑐1 in (ii), We have
𝑥 𝑡 = 8 cos 6𝑡 + 𝑐2 sin 6𝑡
Differentiating (iii) in one times,
𝑥′
𝑡 = −48 sin 6𝑡 + 6𝑐2 cos 6𝑡
Now using the condition 𝑥′ 0 = 0 in (iv),
𝑥′
0 = −48 sin 0 + 6𝑐2 cos 0
⇒ 0 = 6𝑐2
∴ 𝑐2 = 0
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……………….(iii)
.………………(iv)

9. Finally we get, 𝑥 𝑡 = 8cos(6𝑡)
Now we find the period and frequency of the resulting motion.
We know period, 𝑇 =
2𝜋
𝜔
=
2𝜋
6
≈ 1.047 sec 𝑡𝑜 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒/𝑐𝑦𝑐𝑙𝑒
Frequency 𝑓 =
1
𝑇
=
6
2𝜋
=
3
𝜋
≈ 0.955 𝑐𝑦𝑐𝑙𝑒𝑠 𝑎𝑟𝑒 𝑐𝑜𝑚𝑝𝑙𝑒𝑡𝑒𝑑 𝑒𝑣𝑒𝑟𝑦 𝑠𝑒𝑐𝑜𝑛𝑑.
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Department Of Mathematics
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11. References
 Charles R. MacCluer, Industrial Mathematics Modeling in Industry, Science, and Government.
 Google image.
 https://www.youtube.com/watch?v=f2wGE_n5xtA.
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12. THANKS TO ALL
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