Simple Harmonic Motion Periodic Motion: any motion which repeats itself at regular, equal intervals of time.
Link to SHM Applet
Simple Harmonic Motion It is actually anything but simple! Simple harmonic motion is a special case of periodic motion
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
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
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
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
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
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.
What is the equation of this graph? Hint HIVO, HOVIS A cos graph takes 2 radians to complete a cycle
Relation between Linear SHM and Circular Motion Displacement O B O A Time B B A B
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
= 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
= 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
= 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
= 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
= 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
= 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°
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 2 90°
= 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 180°
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 2 270°
= 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 360°
M2 reminder w = Angular velocity = Units rads per sec change in angle time taken = θ t Angular velocity = Units rads per sec
= 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)
If timing begins when x = 0 at the centre of the SHM Displacement O Time B
= 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)
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)
x = 0.5 cos 𝜋 7 = 0.45 m so it is 0.05m from the end A 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 = 0.45 m so it is 0.05m from the end
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
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
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 0.4 2 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 = 0.4 2 x
Some Vehicles Accelerate Starving Vultures Attack Calculus Memory Aid s v a Differentiate Integrate Some Vehicles Accelerate Thanks to Chloe Barnes Starving Vultures Attack
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,
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,
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
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 2 2 +c v 2 2 = boundary conditions : v = 0 when x = A (amplitude) 0 = − ω 2 Amp 2 2 +c c = ω 2 Amp 2 2 v 2 = ω 2 Amp 2 − ω 2 x 2
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
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
Stop and look at Course notes
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
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
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.
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 = 500.2 = 1.414m/s
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
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
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) 0.8 - 20(0.2 + x) 0.8 = 3 x
N1 N2 E 0.8 0.2 T1 x T1 = 0.2 A B T2 20(0.2 − x) 0.8 - 20(0.2 + x) 0.8 = 3 x simplifying -50x = 3 x - 50 3 x = x which is SHM ω 2 = 50 3
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 = - 50 3 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 = wAmp =