Lesson Notes By Weeks and Term v3 - Senior Secondary 2
Simple harmonic motion.
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Simple harmonic motion.
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Subject: Physics
Class: Senior Secondary 2
Term: 1st Term
Week: 5
Theme: Interaction Of Matter, Space And Time
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Students shouldbe able to:Define simpleharmonicmotion. Show the relationshipbetween Linear and angular speed Linearaccelerationand angularacceleration Show the relationshipbetween periodand frequency Calculate the energy in the system explain for cedvibration and resonance.
Consider an object moving in a circle with uniform angular speed. The projection of this uniform circular motion onto a diameter exhibits SH
M. This connection is fundamental to understanding SH
M. Angular Speed ($\omega$): This is the rate of change of angular displacement. It is measured in radians per second (rad/s). $\omega = \frac{\Delta\theta}{\Delta t}$ For one complete revolution (2π radians) in time T (period), $\omega = \frac{2\pi}{T}$ Since $T = 1/f$ (where f is frequency), we also have: $\omega = 2\pi f$ Linear Speed (v): This is the tangential speed of the object along the circular path. It is measured in meters per second (m/s). If an object travels a distance 's' along an arc of radius 'r' subtending an angle '$\theta$' (in radians), then $s = r\theta$. Linear speed $v = \frac{\Delta s}{\Delta t} = \frac{\Delta (r\theta)}{\Delta t} = r \frac{\Delta \theta}{\Delta t}$ Therefore, the relationship is: $v = r\omega$ Where 'r' is the radius of the circular path. In the context of SHM, 'r' corresponds to the amplitude (A) of the oscillation for maximum linear speed. Angular Acceleration ($\alpha$): This is the rate of change of angular speed. $\alpha = \frac{\Delta\omega}{\Delta t}$ (measured in rad/s2).
Linear Acceleration (a): For an object in circular motion, there are two components: Tangential Acceleration ($a_t$): The component tangential to the circle, caused by a change in linear speed. $a_t = r\alpha$. Centripetal (or Radial) Acceleration ($a_c$): The component directed towards the center of the circle, responsible for changing the direction of velocity. $a_c = \frac{v^2}{r} = \frac{(r\omega)^2}{r} = r\omega^2$ In SHM, the acceleration is strictly linear and is directed towards the equilibrium position. The defining equation for SHM is $a = -\omega^2 x$. This 'a' is the linear acceleration of the oscillating particle. The angular frequency '$\omega$' here is characteristic of the SHM itself, derived from the system's properties (e.g., for a mass-spring system, $\omega = \sqrt{k/m}$). It's not the angular acceleration '$\alpha$' of a rotating body, but rather the constant that relates acceleration to displacement in SHM. The maximum linear acceleration occurs at the extreme ends of the oscillation where displacement $x = \pm A$: $a_{max} = \omega^2 A$ Period (T): The time taken for one complete oscillation or cycle. It is measured in seconds (s).
Frequency (f): The number of complete oscillations or cycles per unit time. It is measured in Hertz (Hz), where 1 Hz = 1 oscillation per second. The relationship between period and frequency is inverse: $T = \frac{1}{f}$ or $f = \frac{1}{T}$ Worked Example 1: A simple pendulum completes 20 oscillations in 30 seconds. Calculate its period and frequency.
Solution: Calculate Period (T): $T = \frac{\text{Total time}}{\text{Number of oscillations}} = \frac{30 \text{ s}}{20 \text{ oscillations}} = 1.5 \text{ s}$ Calculate Frequency (f): $f = \frac{1}{T} = \frac{1}{1.5 \text{ s}} = 0.667 \text{ Hz}$ For a system undergoing SHM, the total mechanical energy (sum of kinetic and potential energy) remains constant if there is no damping.
Kinetic Energy (KE): The energy due to motion. $KE = \frac{1}{2}mv^2$ For SHM, the velocity 'v' at any displacement 'x' is given by $v = \omega\sqrt{A^2 - x^2}$, where 'A' is the amplitude. So, $KE = \frac{1}{2}m(\omega\sqrt{A^2 - x^2})^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$ Maximum KE occurs at the equilibrium position (x=0), where $v_{max} = \omega A$: $KE_{max} = \frac{1}{2}m(\omega A)^2 = \frac{1}{2}m\omega^2 A^2$ Potential Energy (PE): The energy stored due to the object's position or state. For a spring-mass system, it's elastic potential energy. The restoring force is $F = -kx$. The spring constant $k = m\omega^2$. $PE = \frac{1}{2}kx^2$ Substituting $k = m\omega^2$: $PE = \frac{1}{2}m\omega^2 x^2$ Maximum PE occurs at the extreme ends of oscillation (x = ±A): $PE_{max} = \frac{1}{2}m\omega^2 A^2$ (This is also the total energy)
Total Mechanical Energy (TE): The sum of KE and PE. $TE = KE + PE$ $TE = \frac{1}{2}m\omega^2(A^2 - x^2) + \frac{1}{2}m\omega^2 x^2$ $TE = \frac{1}{2}m\omega^2 A^2 - \frac{1}{2}m\omega^2 x^2 + \frac{1}{2}m\omega^2 x^2$ $TE = \frac{1}{2}m\omega^2 A^2$ This shows that the total energy is constant and depends only on the mass (m), angular frequency ($\omega$), and amplitude (A) of the oscillation. It is equal to the maximum kinetic energy or maximum potential energy. Since $k = m\omega^2$, we can also write $TE = \frac{1}{2}kA^2$.
Worked Example 2: A 0.5 kg mass oscillates with an amplitude of 0.1 m and an angular frequency of 5 rad/s. Calculate the total mechanical energy of the system.
Solution: Given: m = 0.5 kg, A = 0.1 m, $\omega$ = 5 rad/s. Total Energy $TE = \frac{1}{2}m\omega^2 A^2$ $TE = \frac{1}{2} \times 0.5 \text{ kg} \times (5 \text{ rad/s})^2 \times (0.1 \text{ m})^2$ $TE = \frac{1}{2} \times 0.5 \times 25 \times 0.01$ $TE = 0.5 \times 0.5 \times 25 \times 0.01 = 0.25 \times 25 \times 0.01 = 6.25 \times 0.01 = 0.0625 \text{ J}$ The total mechanical energy is 0.0625 J.
Vehicle Suspension Systems (Automotive Engineering): SHM principles are critical in designing car suspension systems in Nigeria. The springs and shock absorbers in vehicles are engineered to dampen oscillations, preventing the vehicle from bouncing uncontrollably on rough roads common in many parts of the country. A well-designed suspension system attempts to prevent resonance with the natural frequencies of the vehicle chassis, ensuring a smooth ride and vehicle stability, crucial for passenger safety and cargo integrity.
Musical Instruments (Culture & Arts): Many Nigerian musical instruments, like talking drums (Gangan), xylophones (Balafon), and stringed instruments (e.g., Goje), rely on the principles of SHM and resonance. The membranes of drums or strings of instruments vibrate in SHM when struck or plucked, producing sound waves. The design of the instrument's body acts as a resonator, amplifying these vibrations at specific frequencies (its natural frequencies), thus creating the rich, loud musical tones characteristic of Nigerian music. Building and Bridge Design (Civil Engineering): While less visible, the understanding of SHM and resonance is vital for civil engineers designing structures like bridges (e.g., across the River Niger) and high-rise buildings in urban centers like Lagos or Abuja. Structures have natural frequencies. Engineers must design them to avoid these natural frequencies coinciding with potential external driving forces, such as strong winds (e.g., during Harmattan) or seismic vibrations (though less common in Nigeria, it's a general principle). Ignoring resonance can lead to catastrophic structural failure, as demonstrated by the historical Tacoma Narrows Bridge collapse.