Free damped motion with problem analysis

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.
4/4/2019
Department Of Mathematics
University Of Rajshahi
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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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Department Of Mathematics
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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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Department Of Mathematics
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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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Department Of Mathematics
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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
University Of Rajshahi
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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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University Of Rajshhi
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