7.1 Simple Harmonic Motion PDF

7.1 Simple Harmonic Motion Chapter 9 Sections 9.1 – 9.7 7.1.1 Definitions and Equations In simple terms, Simple Harmonic Motion (S.H.M.) is a to and fro motion, e.g. the motion of: 1. a simple pendulum 2. a ma...

7.1 Simple Harmonic Motion Chapter 9 Sections 9.1 – 9.7 7.1.1 Definitions and Equations In simple terms, Simple Harmonic Motion (S.H.M.) is a to and fro motion, e.g. the motion of: 1. a simple pendulum 2. a mass attached to a spring 3. pendulum of a clock 4. a cork floating in sea water 5. atoms in a molecule Consider a mass attached to a spring: 1 2 3 4 5 𝑷. 𝑬. = 𝒎𝒂𝒙 F B 𝑲. 𝑬. = 𝟎 A 0 0 0 0 0 𝑷. 𝑬. = 𝟎 𝑲. 𝑬. = 𝒎𝒂𝒙 A 𝑷. 𝑬. = 𝒎𝒂𝒙 F A 𝑲. 𝑬. = 𝟎 Figure 1 In figure 1, the mass undergoes Simple Harmonic Motion (S.H.M.). In diagram: 1. Mass is in equilibrium. Therefore, O is the equilibrium position or the rest position or the mean position. L. BONELLO 1 7.1 Simple Harmonic Motion 2. When the mass is pulled to position A and released, there is a force called the restoring force which causes the mass to accelerate towards the equilibrium position O. 3. The mass reaches the equilibrium position O. 4. The mass goes past O, slows down, and comes momentarily to rest at B. 5. The mass then repeats the motion, in the opposite direction. Note: 1 oscillation: e.g. from O → A → O → B → O or from A → O → B → O → A or from B → O → A → O → B Observations: 1. The elastic restoring force decreases as the mass approaches O from A because the tension in the spring is proportional to the extension. So, with the mass approaching O, the extension decreases and so the pull exerted by the spring on the mass decreases. Therefore, the resultant force towards O decreases (force is proportional to displacement according to Hooke’s law). Hooke’s Law: 𝐹 = 𝑘 ∆𝑙 𝐹 ∝ ∆𝑙 So, if ∆𝑙 ↓, then 𝐹 ↓ as well. L. BONELLO 2 7.1 Simple Harmonic Motion 2. At the equilibrium position O, the displacement is given as 𝒙 = 𝟎 where: 𝑥 = displacement (m) Displacement means distance in a given direction. So, displacement is a vector quantity. At the two extreme positions A and B, the displacement is given as 𝒙 = 𝑨 where: A = maximum displacement (or amplitude) 3. The mass overshoots the mean position O because the system has inertia and so, has K.E. 4. The mass slows down when it is past O moving towards B. This is because the spring is being compressed. The push of the spring on the mass, which provides the restoring force, is acting downwards towards O. 5. The energy of the vibrating system consists of: (i) Elastic P.E. stored in the spring decreases to zero at the rest position and then increases. P.E. is a maximum when the displacement is a maximum. (ii) K.E. of the moving mass which increases to its maximum value at the mean position and then decreases. K.E. is zero when the displacement is a maximum. L. BONELLO 3 7.1 Simple Harmonic Motion Total energy of system = P.E. + K.E. = constant (No energy losses) 6. The motion of the mass eventually comes to rest, i.e., the oscillations die out because energy is lost from the system since work is done against the frictional forces (due to air resistance which oppose the moving system). This energy appears as an increase in the internal energy of the surroundings as well as of the spring. These oscillations are called damped oscillations and eventually die out. Free oscillations are not hindered by damping and go on forever. 7. At O (mean At A and B position) (extreme ends) ∆l 0 𝑚𝑎𝑥 𝑥 0 𝑚𝑎𝑥 𝐹 0 𝑚𝑎𝑥 𝑎 0 𝑚𝑎𝑥 𝑃. 𝐸. 0 𝑚𝑎𝑥 𝑣 𝑚𝑎𝑥 0 𝐾. 𝐸. 𝑚𝑎𝑥 0 Table 1 L. BONELLO 4 7.1 Simple Harmonic Motion 𝐹 = 𝑘 ∆𝑙 where ∆𝑙 = 𝑥 (extension = displacement) ∴𝐹 ∝𝑥 Also, 𝐹 = 𝑚 𝑎 ∴𝐹 ∝𝑎 1 And 𝐾. 𝐸. = 𝑚 𝑣2 2 ∴ 𝐾. 𝐸. ∝ 𝑣 2 where: ∆𝑙 = Extension (𝑚) 𝑥 = Displacement (𝑚) 𝐹 = Restoring force (𝑁) 𝑎 = Acceleration (𝑚 𝑠 −2 ) 𝑃. 𝐸. = Potential energy (𝐽) 𝑣 = Velocity (𝑚 𝑠 −1 ) 𝐾. 𝐸. = Kinetic energy (𝐽) Conclusions: 1. Acceleration and displacement are always in opposite directions. 2. Acceleration is proportional to the displacement. L. BONELLO 5 7.1 Simple Harmonic Motion Definition of S.H.M. If the acceleration of a body is directly proportional to its distance from a fixed point (displacement from the mean position) 𝑎 ∝ −and 𝑥 is always directed towards that point (equilibrium position), then the motion is simple harmonic. 𝑎 = −(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝑥 𝑎 = − 𝜔2 𝑥 Equation for S.H.M. where: 𝑎 = Acceleration (𝑚 𝑠 −2 ) 𝑥 = Displacement from the equilibrium position (𝑚) 𝜔2 = Positive constant (𝑠 −2 ) 𝑎 Since 𝜔2 = 𝑥 𝑚 𝑠 −2 ∴ Units of 𝜔2 = = 𝑠 −2 𝑚 𝜔 = Angular frequency (𝑠 −1 ) Negative sign − implies that 𝑎 is acting in the opposite direction to 𝑥. Also, 𝐹 = 𝑚 𝑎 𝐹 = − 𝑚 𝜔2 𝑥 since 𝑎 = − 𝜔2 𝑥 𝐹 = −(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡) 𝑥 L. BONELLO 6 7.1 Simple Harmonic Motion 𝐹 ∝ −𝑥 i.e. the restoring force is: 1. directly proportional to the displacement; and 2. acting in the opposite direction to the displacement. where: 𝐹 = Restoring force (𝑁) 𝑚 = Mass (𝑘𝑔) 𝑎 = Acceleration (𝑚 𝑠 −2 ) 𝑥 = Displacement (𝑚) 𝜔2 = Positive constant (𝑠 −2 ) 𝜔 = Angular frequency (𝑠 −1 ) Negative sign − implies that 𝐹 is acting in the opposite direction to 𝑥. So, one can also state that the resultant force acting on a particle performing simple harmonic motion is the restoring force. L. BONELLO 7 7.1 Simple Harmonic Motion Definitions Equilibrium (or rest or mean) position (𝑥 = 0) : This is the position considered when the system is at equilibrium. Restoring force: The restoring force is the force which is experienced by a system undergoing S.H.M. This force causes the system to accelerate towards the equilibrium position. Amplitude 𝐴: The amplitude of an oscillation is the maximum displacement from the mean position. Periodic Time or Period 𝑇: The periodic time or period is the time taken for one oscillation. 2𝜋 …Equation 1 𝑇= 𝜔 where: 𝑇 = Periodic time or period (𝑠) 𝜔 = Angular frequency (𝑠 −1 ) L. BONELLO 8 7.1 Simple Harmonic Motion Frequency 𝑓: The frequency is the number of oscillations per second. 1 𝑓= 𝑇 …Equation 2 where: 𝑓 = Frequency (𝐻𝑧 𝑜𝑟 𝑠 −1 ) 𝑇 = Periodic time or period (𝑠) Substituting Equation 1 in Equation 2: 1 𝑓= 𝑇 1 𝑓= 2𝜋 𝜔 𝜔 𝑓= 2𝜋 𝜔 =2𝜋𝑓 where: 𝜔 = Angular frequency (𝑠 −1 ) 𝑓 = Frequency (𝐻𝑧 𝑜𝑟 𝑠 −1 ) L. BONELLO 9 7.1 Simple Harmonic Motion E.g., A body performs simple harmonic oscillations and has an acceleration of 4 𝑚 𝑠 −2 when displaced by 4 𝑐𝑚 from its equilibrium position. What is the body’s acceleration when it is at its maximum displacement of 7 𝑐𝑚? 𝒙 A Figure 2 𝑎 = − 𝜔2 𝑥 4 = − 𝜔2. 0.04 4 𝜔2 = (ignoring the minus sign) 0.04 𝜔2 = 100 𝑠 −2 𝑎𝑚𝑎𝑥 = − 𝜔2 𝐴 = −100. 0.07 = −7 𝑚 𝑠 −2 (Negative sign → 𝑎 acts in opposite direction to 𝑥) L. BONELLO 10 7.1 Simple Harmonic Motion Energy of an Oscillator 𝑃𝐸 = 𝑚𝑎𝑥 𝐾𝐸 = 0 Extreme End B 𝑃𝐸 = 0 O 𝐾𝐸 = 𝑚𝑎𝑥 Equilibrium Position 𝑃𝐸 = 𝑚𝑎𝑥 A 𝐾𝐸 = 0 Extreme End Figure 3 During an oscillation, there is continuous interconversion of 𝑃𝐸 (maximum at the ends of the oscillation) to 𝐾𝐸 (maximum at the centre of the oscillation) and vice versa. At any instant during the motion, the energy is the sum of the 𝐾𝐸 of the moving mass and the 𝑃𝐸 stored in the system. Assuming no energy losses due to friction or other resistances to motion: 𝐸𝑇 = 𝐾𝐸 + 𝑃𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 At the mean position, all energy is 𝐾𝐸. At the ends of the oscillation, all energy is 𝑃𝐸. In between, energy is 𝐾𝐸 and 𝑃𝐸. Also, provided that there are no energy losses: L. BONELLO 11 7.1 Simple Harmonic Motion 𝐸𝑇 = 𝐾𝐸𝑚𝑎𝑥 = 𝑃𝐸𝑚𝑎𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Graphical Representation of 𝑷𝑬, 𝑲𝑬 and 𝑬𝑻 against Displacement Energy/𝐽 𝐸𝑇 = 𝐾𝐸𝑚𝑎𝑥 = 𝑃𝐸𝑚𝑎𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 O A B 𝑃𝐸 = 𝐾𝐸 − 𝐴 0 +𝐴 Displacement/𝑚 Extreme End Mean Position Extreme End Graph 1 At O At A and B (mean position) (extreme ends) PE 𝟎 𝒎𝒂𝒙 KE 𝒎𝒂𝒙 𝟎 Table 2 L. BONELLO 12 7.1 Simple Harmonic Motion Assuming no energy losses: 𝐸𝑇 = 𝐾𝐸 + 𝑃𝐸 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝐸𝑇 = 𝐾𝐸𝑚𝑎𝑥 = 𝑃𝐸𝑚𝑎𝑥 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 7.1.2 Acceleration- Displacement Graph Graph of 𝒂 / 𝒎 𝒔−𝟐 on the y-axis against 𝒙 / 𝒎 on the x- axis 𝑎 / 𝑚 𝑠 −2 + max - 0 + 𝑥/𝑚 max max max - B O A Graph 2 𝑥/𝑚 𝑎 / 𝑚 𝑠 −2 O 0 0 (mean position) A and B 𝑚𝑎𝑥 𝑚𝑎𝑥 (extreme ends) Table 3 L. BONELLO 13 7.1 Simple Harmonic Motion Looking back at graph 2 on page 13: 𝑎 = − 𝜔2 𝑥 𝑦=− 𝑚 𝑥 The negative sign shows that the slope is negative. ∴ 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 𝑚 = 𝜔2 ∴ 𝜔 = √𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 where: 𝜔 = Angular frequency (𝑠 −1 ) Hence, to find the periodic time from this graph: Once 𝜔 is found from the graph, then: 2𝜋 𝑇= 𝜔 Substituting for 𝜔, 𝑇 is found. In the case of a mass on a light spring, the periodic time is given as: 𝑚 𝑇 = 2𝜋√ 𝑘 where: 𝑇 = Periodic time (𝑠) 𝑚 = mass of oscillating system (𝑘𝑔) 𝑘 = Stiffness constant (or Force constant) (𝑁 𝑚−1 ) L. BONELLO 14