1. Simple Harmonic Motion
Periodic Motion:
any motion which repeats itself at regular, equal intervals of time.

2. Link to SHM Applet

3. Simple Harmonic Motion
It is actually anything but simple!
Simple harmonic motion is a special case of periodic motion

4. Characteristics of simple harmonic motion:
the displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude),
Velocity is maximum when displacement is zero,
Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,
The period does not depend on the amplitude.
It is periodic oscillatory motion about a central equilibrium point,
Equilibrium position
Positive amplitude
Negative amplitude A O B

5. Characteristics of simple harmonic motion:
It is periodic oscillatory motion about a central equilibrium point,
The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude),
Velocity is maximum when displacement is zero,
Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,
The period does not depend on the amplitude. A A
Displacement O
Equilibrium position
Positive amplitude
Negative amplitude A O B
Time B B

6. Characteristics of simple harmonic motion:
It is periodic oscillatory motion about a central equilibrium point,
The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude),
Velocity is maximum when displacement is zero,(equilibrium point)
Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,
The period does not depend on the amplitude. A A
Displacement O
Time B B
Velocity
Time A O B
remember velocity is the gradient of displacement
Positive amplitude A
Equilibrium position O
Negative amplitude B

7. Characteristics of simple harmonic motion:
It is periodic oscillatory motion about a central equilibrium point,
The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude),
Velocity is maximum when displacement is zero,
Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,
The period does not depend on the amplitude. A A
Displacement O
Equilibrium position
Positive amplitude
Negative amplitude A O B
Time B B
Velocity
Time A O B
remember acceleration is the gradient of velocity
Acceleration
Time A O B

8. Characteristics of simple harmonic motion:
It is periodic oscillatory motion about a central equilibrium point,
The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude),
Velocity is maximum when displacement is zero,
Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,
The period does not depend
on the amplitude.
Displacement
Equilibrium position
Positive amplitude
Negative amplitude A O B
Velocity
Time A O B
Acceleration
Time A O B

9. Simple Harmonic Motion
Equilibrium: the position at which no net force acts on the particle.
Displacement: The distance of the particle from its equilibrium position. Usually denoted as x(t) with x=0 as the equilibrium position.
Amplitude: the maximum value of the displacement with out regard to sign. Denoted as xmax or A.

10. What is the equation of this graph?
Hint
HIVO, HOVIS
A cos graph takes 2 radians to complete a cycle

11. Relation between Linear SHM and Circular Motion
Displacement O B O A
Time B B A B

12. Relation between Linear SHM and Circular Motion
Displacement O B O A
Time B B
Velocity
Time A O B A B
Acceleration
Time A O B

13. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A xo xo A A
Displacement O
Time A B B B
At A, t = 0
the displacement is maximum positive x = xo

14. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A xo A A
Displacement O
Time A B B B
At O, t = 𝑇 4
the displacement is zero x = 0

15. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A
-xo xo A A
Displacement O
Time A B B B
At B, t = 𝑇 2
the displacement is maximum negative x = -xo

16. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A xo A A
Displacement O
Time A B B B
At O, t = 3𝑇 4
the displacement is zero x = 0

17. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A xo A xo A
Displacement O
Time A B B B
At A, t = T
the displacement is maximum positive x = xo

18. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A xo xo A A
Displacement O
Time A B B B
At A, t=0
the displacement is maximum positive x = xo
θ = 0°

19. the displacement is zero x = 0 θ = 𝜋 2 90°B O A xo A A
Displacement O
Time A B B B
At O, t = 𝑇 4
the displacement is zero x = 0
θ = 𝜋 °

20. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle B O A
-xo xo A A
Displacement O
Time A B B B
At B, t = 𝑇 2
the displacement is maximum negative x = -xo
θ = π °

21. the displacement is zero x = 0 θ = 3𝜋 2 270°B O A xo A A
Displacement O
Time A B B B
At O, t = 3𝑇 4
the displacement is zero x = 0
θ = 3𝜋 °

22. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle A O B xo A xo A
Displacement O
Time B A B B
At A, t = T
the displacement is maximum positive x = xo
θ = 2π °

23. M2 reminder w = Angular velocity = Units rads per sec
change in angle time taken = θ t
Angular velocity =
Units
rads per sec

24. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle A O M B x xo A A
Displacement O
Time θ B A B B x
At a random point M , the displacement is x
and θ = θ°
since ω = θ t 
θ = ω t
x xo
cos (ω t) = x xo
Using trigonometry
cos θ = 
x = xo cos (ω t)

25. If timing begins when x = 0 at the centre of the SHM
Displacement O
Time B

26. = the time to complete 1 cycle
T = periodic time
= the time to complete 1 cycle A O M B x xo A A
Displacement O
Time θ B A B B x
If however timing begins when x = 0 at the centre of the SHM then the displacement equation is :
ω = θ t 
θ = ω t
x = xo sin (θ)
So x = xo sin (ω t)

27. Displacement Formulae
If x = x0 when t = 0 then use x = xo cos (ω t)
This is usually written as x = A cos (ω t)
where A is the amplitude of the motion
If x = 0 when t = 0 then use x = xo sin (ω t)
This is usually written as x = A sin (ω t)
where A is the amplitude of the motion
Memory aid
Sin graphs start at the origin(centre)
Cos graphs start at the top(end)

28. x = 0.5 cos 𝜋 7 = 0.45 m so it is 0.05m from the endA O M B x
Amp A A
Displacement
0.45
𝜋 7 O
Time
𝜋 7 A B B B x
What is the displacement of a ball from the end undergoing SHM between points A and B starting at A where the distance between them is 1 m and the angle is 𝜋 7 rads
x = A cos (ω t) when t = 0 A =
0.5 i.e the Amplitude
x = 0.5 cos 𝜋 7 = m so it is 0.05m from the end

29. Periodic Time
ω = θ t So t = θ ω
To complete one cycle t = T and q = 2p
T = 2π ω
Tom is 2 pies over weight
So the time to complete one cycle is T = 2π ω
ω = 2π T

30. To find ω use the fact that ω = θ t T= 𝟐𝝅 
Displacement
Time A B O
Amp
What is the displacement of a ball from the centre undergoing SHM between points A and B starting at the A, after 2secs where the distance between them is 1 m and the periodic time is 6
-0.25 2
x = A cos (ω t) when t = 0 A =
0.5 i.e the Amplitude
x = 0.5 cos 2ω
Memory aid
Tom is 2 pies overweight
To find ω use the fact that ω = θ t
T= 𝟐𝝅 
To complete one whole cycle then q = 2p and t = 6
So ω = θ t = 2𝜋 6 = 𝜋 3 M B O A
x = 0.5 cos 2ω = 0.5 cos 2 𝜋 3 = -0.25 x
-0.25

31. To find ω use the fact that ω = θ T T= 𝟐𝝅 
What is the displacement from the of a ball going through SHM between points A and B starting at the centre, after 3secs where the distance between them is 0.8m and the periodic time is 8s A xo 3
Displacement O B B
x = A sin (ω t) when t = 0 A =
0.4 i.e the Amplitude
x = 0.4 sin 3ω
Memory aid
Tom is 2 pies overweight
To find ω use the fact that ω = θ T
T= 𝟐𝝅 
To complete one whole cycle then q = 2p and t = 8
So ω = θ T = 2𝜋 8 = 𝜋 4 M B O A
x = 0.4 sin 3ω = 0.4 sin 3 𝜋 4 = x

32. Some Vehicles Accelerate Starving Vultures Attack
Calculus Memory Aid s v a
Differentiate
Integrate
Some Vehicles Accelerate
Thanks to Chloe Barnes
Starving Vultures Attack

33. dx dt Velocity Equation. x = A cos (ω t) since v = v = -ω A sin (ω t)
Displacement
dx dt
since v =
Velocity
Time A O B
v =
-ω A sin (ω t)
Characteristics of simple harmonic motion:
Velocity is maximum when displacement is zero,

34. dv dt Acceleration Equation. v = -ω A sin (ω t) since a = a =
Velocity
Time A O B
v = -ω A sin (ω t)
dv dt
since a =
a =
-ω2 A cos (ω t)
Acceleration
Time A O B
a =
-ω2 x
since x = A cos (ω t)
Characteristics of simple harmonic motion:
Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,

35. Equation for velocity where v depends on x rather than t
v = -ω Asin (ω t) v depends on t
dx dt = x
Velocity = rate of change of displacement =
dv dt = v = x
Acceleration = rate of change of velocity =
Chain Rule for acceleration in terms of displacement
So acceleration = dv dt =v dv dx

36. vdv= − ω 2 x dx − ω 2 x 2 2 +c v 2 2 = v 2 = ω 2 Amp 2 − ω 2 x 2
Equation for velocity where v depends on x rather than t
v = -ω A sin (ω t) v depends on t
vdv= − ω 2 x dx
Separate the variables
− ω 2 x c
v 2 2 =
boundary conditions : v = 0 when x = A (amplitude)
0 = − ω 2 Amp c
c = ω 2 Amp 2 2
v 2 = ω 2 Amp 2 − ω 2 x 2

37. Equation for velocity where v depends on x rather than t
v 2 = ω 2 Amp 2 − ω 2 x 2
v 2 = ω 2 (Amp 2 −x 2 )
The max velocity occurs at the centre where x = 0
Vmax = w Amp

38. x = A cos (ωt) if x = A when t = 0
Summary
x = A cos (ωt) if x = A when t = 0
x = A sin (ωt) if x = 0 when t = 0
v = -ω A sin (ωt) if x = A when t = 0
Differentiate the displacement equation
𝐯 𝟐 = 𝛚 𝟐 (𝐀 𝟐 −x 𝟐 )
Vmax = ω A
a = -ω2 x
T = 𝟐𝛑 𝛚 Tom is 2 pies over weight

39. Stop and look at Course notes

40. T = for an elastic string
Horizontal elastic string
Consider a GENERAL point P where the extension is x .
Resolving 0 – T = m x but
– = m x so 𝑥 =
Comparing with the S.H.M eqn x = –w2x w2 = T
+ve x x l
T = for an elastic string

41. Bob pulled so that extension = 0.2m. Find time for which string taught
Ex.1 String l = 0.8m l = 20N m = 0.5kg
Bob pulled so that extension = 0.2m.
Find time for which string taught
b) Time for 1 oscillation
c) Velocity when ext. = 0.1m T
+ve x x l

42. Horizontal elastic string
Resolving 0 – T = m x but T = for an elastic string
– = 0.5 x so x =
Comparing with the S.H.M eqn x = –w2x
l = 0.8m l = 20N m = 0.5kg T
+ve x x l
w2 = 50
Hence w = 50
As the string goes slack at the natural length it then travels with constant velocity before the string goes taught again.

43. a)
This is the periodic time when the string is tight
b) Time for which initially taught is a T so t =
As the initial extension is 0.2 m then
Amp = 0.2m
Velocity when string goes slack v = w Amp as x = 0
vslack = 500.2 = 1.414m/s

44. So the distance travelled when string is slack is 4  0.8
So time for which slack
So total time for 1 oscillation = = 3.15secs
c) Vel. when ext = 0.1
v2 = w2(Amp2 – x2)
v2 = 50(0.22 – 0.12) = 1.5ms–1 v = 1.225ms–1

45. Two Connected Horizontal Elastic Strings
A and B are two fixed points on a smooth surface with AB = 2m. A string of length 1.6m and modulus 20N is stretched between A and B. A mass of 3kg is attached to a point  way along the string from A and pulled sideways 9cm towards A. Show that the motion is S.H.M and find the max acceleration and max velocity. A B

46. N1 N2 E
0.8
0.2 T1 x
T1 =
0.2 A B T2
As x is increasing towards A make  +ve x is measured from the equilibrium position
+ve direction  T1 – T2 = 3a
20(0.2 − x) (0.2 + x) 0.8 = 3 x

47. N1 N2 E
0.8
0.2 T1 x
T1 =
0.2 A B T2
20(0.2 − x) (0.2 + x) = 3 x
simplifying
-50x = 3 x
x = x
which is SHM
ω 2 = 50 3

48. Max acc occurs at the max extension i.e Amp = 0.09
0.8
0.2 T1 x
T1 =
0.2 A B T2
x = x
ω 2 = 50 3
Max acc occurs at the max extension i.e Amp = 0.09
Max acc =
50 3
× 0.09 = 0.37ms-1
Max velocity = wAmp =