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

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. 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 Time B B 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 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 Time B B Velocity Time A O B Acceleration Time A O B Positive amplitude A remember acceleration is the gradient of velocity 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 Time B B A A Displacement O Positive amplitude A Time Equilibrium position O Negative amplitude B B B

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Questions 1 1.A body performs simple harmonic motion when its acceleration is proportional to its displacement from a fixed point and is directed towards that point. What additional condition must be satisfied for the oscillations of a pendulum to be simple harmonic? ............................................................................................................................ The diagram shows a long pendulum which is oscillating between points A and B. The pendulum takes 5.72 s to swing from A to B. Calculate its period T. T = ........................................................ A B

Questions 2 2. A body performs simple harmonic oscillations. The graph shows how the acceleration a of the body varies with time t. State the frequency of the oscillations. ........................................................... .......................................................... Add to the graph above a curve showing how the velocity of the same body varies with time over the same period. On the same graph, sketch a graph to show how the force F acting on the same body varies with time over the same period.

Questions 2 answer 2. A body performs simple harmonic oscillations. The graph shows how the acceleration a of the body varies with time t. State the frequency of the oscillations. Frequency = 0.77 Hz Add to the graph above a curve showing how the velocity of the same body varies with time over the same period. On the same graph, sketch a graph to show how the force F acting on the same body varies with time over the same period.

Questions 4 4. A body oscillates with simple harmonic motion. On the axes below sketch a graph to show how the acceleration of the body varies with its displacement. a/ ms-2 X / m

Questions 4 4. A body oscillates with simple harmonic motion. On the axes below sketch a graph to show how the acceleration of the body varies with its displacement. a/ ms-2 X / m

A A Displacement O Time B B Velocity Time A O B Acceleration Time A O B

A A Displacement O Time B B

the displacement is maximum positive x = xo B xo xo A A Displacement O Time B A B B At A, t=0 the displacement is maximum positive x = xo

the displacement is zero x = 0 B xo A A Displacement O Time B A B B At O, t = T/4 the displacement is zero x = 0

the displacement is maximum negative x = -xo B -xo xo A A Displacement O Time B A B B At B, t = T/2 the displacement is maximum negative x = -xo

the displacement is zero x = 0 B xo A A Displacement O Time B A B B At O, t = 3T/4 the displacement is zero x = 0

the displacement is maximum positive x = xo B xo A xo A Displacement O Time B A B B At A, t = T the displacement is maximum positive x = xo

the displacement is maximum positive x = xo B xo xo A A Displacement O Time B A B B At A, t=0 the displacement is maximum positive x = xo θ = 0°

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

the displacement is maximum negative x = -xo B -xo xo A A Displacement O Time B A B B At B, t = T/2 the displacement is maximum negative x = -xo θ = π 180°

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

the displacement is maximum positive x = xo 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°

θ At a random point M , the displacement is x θ = θ° since ω = θ /t  B x xo A A Displacement O Time θ B A B B x At a random point M , the displacement is x θ = θ° since ω = θ /t  θ = ω t Using trigonometry cos θ = x / xo  cos (ω t) = x / xo x = xo cos (ω t)

try it out when the angle is π/2 , π, 3π/2, and 2 π M O B x xo A A Displacement O Time π/ 7 B A B B x What is the displacement of a ball going through SHM between points A and B where the distance between them is 1 m and the angle is π/ 7 x = xo cos (ω t) or x = xo cos (θ) x = 0.5 cos (π/ 7) = 0.45 m try it out when the angle is π/2 , π, 3π/2, and 2 π

That is the displacement sorted out That is the displacement sorted out. How about the velocity and acceleration? since v = dx/dt Velocity Time A O B v = -ω xo sin (ω t) since a = dv/dt a = -ω2 xo cos (ω t) since x = xo cos (ω t) Acceleration Time A O B a = -ω2 x

a = -ω2 x acceleration is proportional to the displacement but in the opposite direction -ω2 is the gradient of the graph a / ms-2 a / ms-2 x / m x / m large value of fast moving object small value of ω slow moving object

Phase angles In phase

Phase angles In phase Out of phase A phase difference of π

Phase angles In phase Out of phase A phase difference of π Out of phase A phase difference of π/2

x = xo cos (ω t) v = -ω xo sin (ω t) a = -ω2 x

For a mass-spring system F = - kx where F is the restoring force (the force exerted by the spring) k is the spring constant, the bigger the value of k, the stiffer the spring. x is the extension in the spring due to the mass hanging from it. The negative sign means that the extension is in the opposite direction to the force x F mg

A force is applied ( to the right) to displace the trolley a displacement x, the trolley accelerates to the left due to the effect of the restoring force of the spring. F = - kx since F = ma and a = -ω2 x ma = - kx a = - kx/m -ω2 x = - kx/m ω2 = k/m x restoring Force Force causing the extension

k ω2 = m k ω = m 2 π since ω = T 2 π k ω = = m T m T = 2 π k the bigger the mass, the bigger the period (slower oscillations) the bigger the k value (the stiffer the spring) the smaller the period (faster oscillations)

k ω2 = m k ω = m 2 π f since ω = k = 2 π f ω = m 1 k f = 2 π m the bigger the k value (the stiffer the spring), the higher the frequency (faster oscillations). the bigger the mass, the smaller the frequency (slower oscillations).

Measure the mass m of the trolley Calculate the k value of the spring Tether the trolley between two springs Use a newtonmeter to measure the force needed to displace the trolley a measured displacement from the equilibrium point Set the trolley to oscillate and measure the period compare your experimental value of T with the value from the equation m T = 2 π k x restoring Force Force causing the extension

Force and extension Hooke’s Law: Extension is proportional to the applied force that caused it. When you apply a force to a spring, you fix the spring at one end and suspend a sequence of masses from the other. Then take measurements of the extension of the spring as you change the masses, altering the applied force. extension/m force/N k= force extension beyond the elastic limit Usually, we plot the quantity that we are varying along the x axis and what we measure up the y axis. But not here. In this case, the quantity we vary, the force, is plotted on the y axis, and the length we measure is on the x axis. The important result is that the graph is a straight line, provided we keep the applied force reasonably small (the spring goes beyond its elastic limit if it deforms). The straight line shows that the spring extends by equal amounts for equal masses added. And the gradient is the spring constant k. F = k x

Force and extension Hooke’s Law: Extension is proportional to the applied force that caused it. When the mass applies a downward force on the spring, the spring applies an upward force on the mass. This is called the ‘restoring force’. Both the applied force and the restoring force have the same magnitude but have opposite directions. x F applied F restoring F applied = k x F restoring = - k x In simple harmonic motion, we are interested in the restoring force.

Springs in series Springs can be combined to carry a single load. The effective spring constant k will depend on how these springs are arranged. In here they are arranged in series. The load is suspended from the lower spring but the force F acts on both springs. k k kseries = F/2x kseries = kx/2x F applied 2x kseries = k/2 two springs in series is equivalent to a spring that is half the stiffness

Springs in parallel In here they are arranged in parallel. The force is now shared between the two springs, so each spring has a force of F/2. the extension for each spring is F/2k k k x/2 F applied kparallel = F/(F/2k) kparallel = 2k two springs in parallel is equivalent to a spring that is twice the stiffness

k ω = m m T = 2 π k Investigate the relation between k and T Vary k by using different spring combinations measure T Plot 1/k½ against T or plot 1/k against T2 Find the gradient

How to calculate Elastic potential energy EPE When you apply a force on a spring, you do work on it (transfer energy to it). This work is going to be stored in the spring until it is released in the form of Elastic Potential Energy EPE. Work done = EPE Since Work done = Force x distance But the force that is applied to stretch the spring a certain distance (extension) is not uniform, it changes from zero to a maximum. extension / m Force / N x F Work done = average Force x distance gradient = k average Force = (maximum force + 0)/2 Work done = ½ Force x distance = ½ Fx Work done = the area under the curve since F = k x Work done = ½ k x2 EPE = ½ k x2

Energy of an oscillator Displacement Time A B O xo A O B Velocity Time A O B x +xo -xo TE KE Time A O B KE GPE Time A O B GPE

A spring-mass system undergoing SHM position A, KE = 0 GPE = mg(xo+xo) EPE = 0 xo position O (equilibrium), KE = ½ mv2 GPE = mgxo EPE = ½ kxo2 xo position B KE = 0 GPE = 0 EPE = ½ k(xo+xo)2

Chapter 8 Q5 un-stretched spring 0.4 kg mass added causing a 0.06m extension the mass is pulled another 0.06m and released. so it oscillates with SHM

position A, KE = ..……..., GPE = ………, EPE = …………. position O, KE = ………., GPE = …….., EPE = ………… position B, KE = ………., GPE = ………., EPE = ……….. KE = ½ mv2 where v = -ωxo sin(ωt) = -√ (k/m) xo sin(ωt) GPE = mgh EPE = ½ kx2

position A, KE = ..……..., GPE = ………, EPE = …………. max position A, KE = ..……..., GPE = ………, EPE = …………. GPE = 0.4x9.8x0.12 = 0.47 J max medium medium position O, KE = ………., GPE = …….., EPE = ………… vmax = -(√ (k/m)) xo = -(√ (65.3/0.4)) x 0.06 = 0.77ms-1 KE = ½ x 0.4 x (0.77)2 = 0.118 J GPE = 0.4x9.8x0.06 = 0.235 J EPE = ½ x 65.3 x ( 0.06)2 = 0.118 J max position B, KE = ………., GPE = ………., EPE = ……….. k = F/x = 0.4x9.8/0.06 = 65.3 Nm-1 EPE = ½ x 65.3 x ( 0.12)2 = 0.47 J KE = ½ mv2 where v = -ωxo sin(ωt) = -(√ (k/m) )xo sin(ωt) GPE = mgh EPE = ½ kx2