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

Characteristics of simple harmonic motion: 1.It is periodic oscillatory motion about a central equilibrium point, 2.the displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), 3.Velocity is maximum when displacement is zero, 4.Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, 5.The period does not depend on the amplitude. Equilibrium position Positive amplitude Negative amplitude A O B

Characteristics of simple harmonic motion: 1.It is periodic oscillatory motion about a central equilibrium point, 2.The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), 3.Velocity is maximum when displacement is zero, 4.Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, 5.The period does not depend on the amplitude. Displacement Time AA BB O Equilibrium position Positive amplitude Negative amplitude A O B

Characteristics of simple harmonic motion: 1.It is periodic oscillatory motion about a central equilibrium point, 2.The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), 3.Velocity is maximum when displacement is zero,( equilibrium point) 4.Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, 5.The period does not depend on the amplitude. Displacement Time AA BB O Velocity Time A O B O remember velocity is the gradient of displacement Equilibrium position Positive amplitude Negative amplitude A O B

Characteristics of simple harmonic motion: 1.It is periodic oscillatory motion about a central equilibrium point, 2.The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), 3.Velocity is maximum when displacement is zero, 4.Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, 5.The period does not depend on the amplitude. Displacement Time AA BB O Velocity Time A O B O Acceleration Time A O B remember acceleration is the gradient of velocity Equilibrium position Positive amplitude Negative amplitude A O B

Characteristics of simple harmonic motion: 1.It is periodic oscillatory motion about a central equilibrium point, 2.The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), 3.Velocity is maximum when displacement is zero, 4.Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, 5.The period does not depend on the amplitude. Displacement Time AA BB O Displacement Time AA BB O Equilibrium position Positive amplitude Negative amplitude A O B

Link to SHM Applet

Displacement Time AA BB O Velocity Time A O B O Acceleration Time A O B Relation between Linear SHM and Circular Motion

Displacement Time AA BB O

Displacement Time AA BB O At A, t=0 the displacement is maximum positive x = x o A A xoxo O B B xoxo

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

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

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

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

Displacement Time AA BB O At A, t=0 the displacement is maximum positive x = x o θ = 0° A A xoxo O B B xoxo

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

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

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

Displacement Time A A B B O At A, t = T the displacement is maximum positive x = x o θ = 2π 360° A A xoxo O B B xoxo

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

Displacement Time AA BB O 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 A A xoxo O B B x x M x = x o cos ( θ ) π/ 7 or x = x o cos (ω t) x = 0.5 cos ( π/ 7 ) = 0.45 m try it out when the angle is π/2, π, 3π/2, and 2 π

since x = x o cos (ω t) That is the displacement sorted out. How about the velocity and acceleration? a = -ω 2 x o cos (ω t) since v = dx/dt since a = dv/dt v = -ω x o sin (ω t) a = -ω 2 x Velocity Time A O B O Acceleration Time A O B

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

x = x o cos (ω t) v = -ω x o sin (ω t) a = -ω 2 x

For a mass-spring system x where F is the restoring force (the force exerted by the spring) is the modulus of elasticity, the bigger the value of, 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. F mg

and a = -ω 2 x 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 x Force causing the extension restoring Force since F = ma ma = - kx a = - kx/m -ω 2 x = - kx/m ω 2 = k/m

ω 2 = since ω = k m ω = k m 2 π T ω = k m 2 π T = m k T = 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)

ω 2 = since ω = k m ω = k m 2 π f ω = k m 2 π f = k m 2 π f = 1 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).

x Force causing the extension restoring Force 1.Measure the mass m of the trolley 2.Calculate the k value of the spring 3.Tether the trolley between two springs 4.Use a newtonmeter to measure the force needed to displace the trolley a measured displacement from the equilibrium point 5.Set the trolley to oscillate and measure the period 6.compare your experimental value of T with the value from the equation m k 2 π T =

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. 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 extension/m force/N k= force extension beyond the elastic limit

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. F applied = k x F restoring = - k x x F applied F restoring In simple harmonic motion, we are interested in the restoring force.

Springs in series 2x F applied 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 series = F/2x k k k series = kx/2x k series = k/2 two springs in series is equivalent to a spring that is half the stiffness

Springs in parallel x/2 F applied 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 parallel = F/(F/2k) k k k parallel = 2k two springs in parallel is equivalent to a spring that is twice the stiffness

m k 2 π T = ω = k m Investigate the relation between k and T Vary k by using different spring combinations measure T Plot 1/k ½ against T or plot 1/k against T 2 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. Work done = average Force x distance 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 x 2 EPE = ½ k x 2 extension / m Force / N x F gradient = k

Velocity Time A O B O A O B A Displacement Time AA BB O xoxo OO O A Energy of an oscillator A O B +x o -x o KE Time A O B O A O B A O GPE Time A O B O A O B A O x GPE KE TE

position A, KE = 0 GPE = mg(x o +x o ) EPE = 0 A spring-mass system undergoing SHM position O (equilibrium), KE = ½ mv 2 GPE = mgx o EPE = ½ kx o 2 position B KE = 0 GPE = 0 EPE = ½ k(xo+xo) 2 xoxo xoxo

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 Chapter 8 Q5

position A, KE =..……..., GPE = ………, EPE = …………. position B, KE = ………., GPE = ………., EPE = ……….. position O, KE = ………., GPE = …….., EPE = ………… KE = ½ mv 2 where v = -ωx o sin(ωt) = -√ (k/m) xo sin(ωt) GPE = mgh EPE = ½ kx 2

position A, KE =..……..., GPE = ………, EPE = …………. position B, KE = ………., GPE = ………., EPE = ……….. position O, KE = ………., GPE = …….., EPE = ………… KE = ½ mv 2 where v = -ωx o sin(ωt) = -(√ (k/m) )xo sin(ωt) GPE = mgh EPE = ½ kx max 0 medium max GPE = 0.4x9.8x0.12 = 0.47 J k = F/x = 0.4x9.8/0.06 = 65.3 Nm -1 EPE = ½ x 65.3 x ( 0.12) 2 = 0.47 J v max = -(√ (k/m)) x o = -(√ (65.3/0.4)) x 0.06 = 0.77ms -1 KE = ½ x 0.4 x (0.77) 2 = J GPE = 0.4x9.8x0.06 = J EPE = ½ x 65.3 x ( 0.06) 2 = J