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Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test SIMPLE HARMONIC MOTION KINEMATICS 2 INU0115/515 (MATHS 2) Dr Adrian Jannetta MIMA CMath FRAS Simple Harmonic Motion 1 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Objectives This is the final topic of the kinematics section of the course. We’re going to study about simple harmonic motion — a type of oscillation often encountered in science and engineering. We usually abbreviate simple harmonic motion to SHM in discussions. • In this presentation we will derive the equations of motion for SHM. • We’ll learn how to use the SHM equations to answer questions about the motion. • We’ll see that there is a close link between uniform motion in a circle and SHM. Part of the derivation involves integration using a trig substitution. It would be beneficial to revise that method before continuing. Also, you’ll need to be familiar with compound trig identities. Simple Harmonic Motion 2 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Description of Simple Harmonic Motion Simple Harmonic Motion (SHM) occurs when an elastic force, described by Hooke’s Law, is present. If the force F on an object is proportional to the objects’ displacement x from a point O and is directed towards that point then the particle will move with SHM. F P O x In equation form the force is given by F = −kx, where k is a constant. The equations for SHM can represent a model a range of phenomena; a mass on a spring, motion of a pendulum, molecular vibration. Understanding SHM will give you a better understanding of more complicated types of motion. Simple Harmonic Motion 3 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Simple Harmonic Motion The animation shows a mass on a spring oscillating with simple harmonic motion. The mass starts from rest (v = 0) some distance x from equilibrium. A force acting on the mass towards the centre (and therefore acceleration a too) is a maximum. The mass moves towards the centre gaining speed until it is moving fastest through the equilibrium point. The direction of the force changes on the other side of the equilibrium point. The speed decreases to zero as the force tries to bring the mass back to equilibrium. The motion is reversed and the particle gathers speed again. The restoring force and non zero velocity at equilibrium cause the mass to oscillate indefinitely without friction. Simple Harmonic Motion 4 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test SHM Equation of motion The force on the particle is F = −kx Newton’s 2nd law states that F = ma so that m d2 x = −kx dt 2 ⇒ d2 x k =− x dt 2 m To simplify the mathematics we’ll express the constant of proportionality in the following form: d2 x = −ω2 x dt 2 where ω2 = mk is a positive constant1 . We will try to find a solution for x of the form x = f (t). Although there are methods for solving the second order differential equation we have not studied them on this course. Instead our strategy will be to rewrite the differential equation as a first order equation and solve it by separating the variables. 1 The reason for choosing ω2 rather than ω will become clear later. Simple Harmonic Motion 5 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Solving the SHM equation The particle begins at rest (v = 0) at a distance A from the origin: The SHM equation with boundary conditions is: d2 x = −ω2 x , t = 0 , x = A , v = 0 dt 2 (1) Acceleration is the rate of change of velocity, so dv = −ω2 x dt Using the chain rule we can write the LHS as dv dt = = (2) dx dv × dt dx dv v dx The SHM equation (2) becomes dv = −ω2 x dx We can solve this using the separation of variables method. v Simple Harmonic Motion 6 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Finding the velocity Separating the variables gives: ∫ ∫ v dv = − The solution becomes 1 2 2v 2 ω2 x dx v Integrate both sides 1 2 2v = − 12 ω2 x2 + C1 where C1 is a constant which we calculate using a boundary condition. = − 12 ω2 x2 + 12 ω2 A2 = ω2 A2 − ω2 x2 Taking a common factor of ω2 : v2 = ω2 (A2 − x2 ) Taking a square-root: Since x = A and v = 0 (at t = 0) then C1 = 12 ω2 A2 . p v = ±ω A2 − x2 The ± means that there is a positive and negative velocity associated with each value of x. This is because the particle can pass through each point in two directions! Simple Harmonic Motion 7 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Solving the SHM equation p The velocity of the particle was found to be v = ±ω A2 − x2 . We must integrate the velocity to obtain a solution for x. Taking the positive root and writing the velocity as a derivative: p dx = ω A2 − x 2 dt Separate the variables and integrate: ∫ ∫ 1 dx = ω dt p A2 − x2 Use the substitution x = A sin θ in the LHS: ∫ (A cos θ dθ ) = p A2 − A2 sin2 θ ∫ ωt dθ = ωt θ = ωt + C Simple Harmonic Motion 8 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Solving the SHM equation Since θ = sin−1 x A we have: sin−1 x = ωt + C A Use the other boundary condition; x = A when t = 0 A C = sin−1 = sin−1 (1) = A π 2 C is an angle; it is usually called the phase angle ϕ. So x sin−1 = ωt + π2 A Take sine of both sides and rearrange: x = sin(ωt + π2 ) A The solution to the SHM equation is x = A sin(ωt + π2 ) We can use a compound angle identity to simplify further: x = A cos ωt Simple Harmonic Motion 9 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Displacement-time graph The function x = A cos ωt defines the displacement x from the equilibrium position for the particle at time t. x A x = A cos ωt t −A 2π ω The constant A is the amplitude, ω determines the period of oscillation (or the frequency). Simple Harmonic Motion 10 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Period and angular frequency The amplitude of oscillation (the maximum value that x can take) is A. This occurs at values of t = 0 (the initial displacement) and whenever ωt = 2πn ± π2 , which means whenever time takes the value t= 2πn π ± ω 2ω The period of oscillation T is found by setting ωt = 2π and rearranging to get 2π T= (3) ω where T is measured in seconds. This equation for the period is exactly the same form as the period for circular motion. The constant ω in that context was called angular speed. In the equations of SHM the constant ω is called the angular frequency. Simple Harmonic Motion 11 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Simple Harmonic Motion The equation of motion for a particle moving with simple harmonic motion is d2 x = −16x dt 2 The particle is initially at rest and released 0.05m from its equilibrium position. Find the period of oscillation and particle displacement after 10 s. The angular frequency is ω = 4 rad s−1 . The period of oscillation is T= 2π π = ≈ 1.57 seconds 4 2 The particle is released from rest so we’ll use the simple form of the general solution: x = A cos 4t (4) We know that when t = 0, x = 0.05m: 0.05 = A cos 0 ∴ A = 0.05 So the position of the particle is given by x = 0.05 cos 4t. When t = 10 then x = −0.033 metres. Simple Harmonic Motion 12 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test A particle oscillates with SHM between two points separated by 0.5 metres each second. Calculate the maximum speed and acceleration of the particle. Write down the equation governing the displacement of the particle. The picture below shows the situation: 0.25 m 0.25 m O The amplitude is half of the distance between the points: A = 0.25 m. The angular frequency is related to the period via T = 2π ω= = 2π 1 The maximum speed: 2π ω, so vmax = ωA = 2π × 0.25 = 0.5π ≈ 1.6 m s−1 The maximum acceleration: amax = ω2 A = (2π)2 × 0.25 = π2 ≈ 9.9 m s−2 The displacement equation for this system is: x = 0.25 cos(2πt) Simple Harmonic Motion 13 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test SHM: general case We just saw the simple case of a particle starting from rest and undergoing SHM. If the particle has some non-zero initial velocity v0 the motion will still be SHM but the displacement equation is x = P sin ωt + Q cos ωt (5) The derivation of this result given in the Appendices of your textbook (volume 2). We saw previously that functions of this form are shifted and scaled sine (or cosine) waves. That means we could also represent the displacement as: x = A sin(ωt + ϕ) (6) where A is the amplitude of oscillation and ω is the angular frequency. The values are related to those in 5 through A2 = P 2 + Q 2 and tan ϕ = Q P Also note the following terms associated with SHM solutions: • Angle ωt + ϕ is called the phase. • Angle ϕ is called the phase constant. Simple Harmonic Motion 14 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test SHM: general case The general SHM equation with boundary conditions is: d2 x = −ω2 x , t = 0 , x = x0 , v = v0 dt 2 The solution, which describes the position at time t, is x = A sin(ωt + ϕ) We have studied functions like this in semester 1! They can be expressed as a sum of a sine and a cosine: x = P sin ωt + Q cos ωt , A2 = P 2 + Q2 2π The period of oscillation T is given by T = . The constant ω is called the angular ω frequency of the system. r v2 The constant A represents the amplitude of the oscillation, where A = ω02 + x02 The velocity of the particle is given by p v = ±ω A2 − x2 Maximum velocity occurs when x = 0 so that vmax = ±ωA. The ± indicates that the direction of the velocity depends on which direction the particle is moving through x = 0. The maximum speed is simply ωA. Simple Harmonic Motion 15 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Displacement-time graph (general case) The general solution to the SHM equation is a sine wave. f (t) 2π ω A x = A sin(ωt + ϕ) x0 x = sin ωt t ϕ ω −A ϕ in the graph above is called the phase angle. The phase angle ω determines where in the cycle the motion starts, i.e. at what point in the cycle that t = 0. The angle Simple Harmonic Motion 16 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test A mass on a spring oscillates with SHM according to the equation d2 x = −9x , x(0) = 2 , v(0) = 6 dt 2 (a) Given that the displacement equation has the form x = P sin ωt + Q sin ωt find the values of P and Q. (b) Find the smallest, positive value of t at which the displacement of the mass first reaches a maximum value. (a) First — note that ω = 3. Therefore x = P sin 3t + Q cos 3t Use the initial conditions: x = 2 when t = 0: 2 = P sin 0 + Q cos 0 ∴2=Q And so x = P sin 3t + 2 cos 3t. We’ll have to differentiate to get v: v= dx = 3P cos 3t − 6 sin 3t dt Now use v = 6 when t = 0. 6 = 3P cos 0 − 6 sin 0 6 = 3P ∴2=P The displacement equation is x = 2 sin 3t + 2 cos 3t . Simple Harmonic Motion 17 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test (b) Given that x = 2 sin 3t + 2 cos 3t We’re going to convert it to a shifted, scaled sine wave. Let’s choose the form x = R sin(3t + α) where R and α are constants to be found. By expanding with a compound angle formula we can write: R sin(3t + α) = R [sin 3t cos α + cos 3t sin α] = (R cos α) sin 3t + (R sin α) cos 3t Comparing the RHS to the displacement equation above leads to: R cos α = 2 R sin α = 2 From previous work we know how to find R: R2 = 22 + 22 = 8 Dividing the equations gives tan α = 1 and so α = quadrant. The displacement equation is: x= Simple Harmonic Motion p ∴R= π 4. p 8 The equations suggest α is in the first 8 sin(3t + π4 ) 18 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM Given that x= General case SHM & Circular motion Test p 8 sin(3t + π4 ) we’re looking for the value of t which maximises the value of x. The sine function sin(⋆) reaches a maximum when ⋆ = π2 . In this case we’d have to set 3t + π4 = Solve this to get t = π 2 π 12 . The period of this system is 2π 3 ; That means a maximum was also π 7π reached at t = 12 − 2π = − 3 12 π 3π And the next will take place at t = 12 + 2π 3 = 4 . So with t = π 12 Simple Harmonic Motion — we definitely found the smallest, positive value of t. 19 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test SHM and Circular Motion SHM is closely related to the uniform circular motion that you’ve seen recently. y P P′ r sin θ θ r O x O As the particle P travels around the circle the horizontal projection P ′ will move vertically so that y = r sin θ . Since y = r sin θ satisfies the SHM equation y ′′ = r 2 y then the motion of P ′ is actually SHM. Similarly, the horizontal motion of P ⋆ is also SHM. r cos θ O P⋆ This link means that the equations for period, speed and acceleration that we obtained for circular motion also apply to situations involving SHM! Simple Harmonic Motion 20 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test SHM equations Because of the similarities between SHM and circular motion we can state the following relationships for period, speed and acceleration: The period — time (in seconds) taken to make one revolution — is given by: 2π T= ω where ω is the angular frequency (in rad s−1 ). The linear speed (in m s−1 ) of the particle is: v = ωr The magnitude of the acceleration (m s−2 ) is a = ω2 r Simple Harmonic Motion 21 / 22 Adrian Jannetta Introduction Description of SHM Derivation Analysis of SHM General case SHM & Circular motion Test Test yourself... Let’s finish with a quick assessment of your knowledge. 1 The SHM equation of motion is d2 x = −4x dt 2 Given x(0) = 3 write down expressions for x(t) and v(t). What is the period of this system? 2 A particle oscillates between two points separated by 0.5 m with a period of 2 seconds. Calculate the maximum speed of the particle to 1DP. 3 A particle moves so that it’s position is defined by x = 5 sin(3t + 1) Is this simple harmonic motion? Answers: 1 x = 3 cos 2t and v = −6 sin 2t. Period is 2π 3 seconds. 2 Use vmax = ωA; vmax = 0.79 m s−1 . 3 Differentiate twice: ẍ = −45 sin(3t + 1) = −9 × 5 sin 3t = −9x By showing ẍ = −9x we proved the particle is moving with SHM. Simple Harmonic Motion 22 / 22 Adrian Jannetta