Oscillatory Motion Lecture Notes PDF

Summary

This document provides a detailed overview of oscillatory motion, focusing on concepts such as simple pendulums and mass-spring systems, presented in a lecture format. It includes equations and examples related to calculating various parameters, such as frequency, period and amplitude during oscillatory motion.

Full Transcript

Oscillatory Motion Oscillation is the process of swinging or moving back and forth in a steady, uninterrupted manner, and oscillatory motion is the movement created by this process. Examples of Oscillatory Motion Simple pendulum Mass spring system Tuning fork Swing in a park ...

Oscillatory Motion Oscillation is the process of swinging or moving back and forth in a steady, uninterrupted manner, and oscillatory motion is the movement created by this process. Examples of Oscillatory Motion Simple pendulum Mass spring system Tuning fork Swing in a park Mass spring system The motion of a mass spring system along a straight line can be represented by the projection of uniform circular motion along a diameter of a reference circle. 𝒙 = 𝑨 𝐜𝐨𝐬 𝒘𝒕 + 𝝓 Where: 𝒙 = 𝑨 𝐜𝐨𝐬 𝒘𝒕 + 𝝓 𝒙 is the displacement relative to the equilibrium position. 𝑨 is the amplitude (maximum displacement of the oscillating system). 𝒘 is the angular velocity or angular frequency 𝟐𝝅 𝝎= = 𝟐𝝅𝒇 𝑻 𝑻 is the periodic time (The time required for making one complete cycle or oscillation). 𝒇 is the frequency (The number of oscillations or cycles that happened in one second). 𝒕 is the time. 𝝓 is the phase constant or phase angle. Displacement (𝒙), Velocity (𝒗), and Acceleration (𝒂) Displacement (𝑥) Maximum displacement 𝒙𝒎𝒂𝒙 = ± 𝑨 𝒙 = 𝑨 𝐜𝐨𝐬 𝒘𝒕 + 𝝓 Maximum velocity Velocity (𝑣) 𝒗𝒎𝒂𝒙 = ± 𝑨𝒘 𝒅𝒙 Maximum acceleration 𝒗= = −𝑨𝒘 𝐬𝐢𝐧(𝒘𝒕 + 𝝓) 𝒂𝒎𝒂𝒙 = ± 𝑨𝒘𝟐 𝒅𝒕 Acceleration (𝑎) 𝒅𝒗 𝒂= = −𝑨𝒘𝟐 𝐜𝐨𝐬(𝒘𝒕 + 𝝓) 𝒅𝒕 𝒂 = −𝒘𝟐 𝒙 ⟹ Equation of Simple Harmonic Motion (SHM) Example A body oscillates with Simple Harmonic Motion (SHM) along the 𝒙 – axis. Its displacement is varying with time according to the equation: 𝜋 𝑥 = 𝟒 cos(𝜋 𝑡 + ) 𝟒 where t is in second, 𝒙 in meter and the angle in rad. a) Find the amplitude, frequency and period of the motion. b) What are the velocity and acceleration at 𝑡 = 1 sec? c) What is the maximum velocity of the body? d) What is the maximum acceleration of the body? Solution: 𝝅 ∵ 𝒙 = 𝟒 𝒄𝒐𝒔(𝝅 𝒕 + ) 𝟒 ∵ 𝒙 = 𝑨 cos(𝒘𝒕 + 𝝓) 𝝅 ∴ 𝑨 = 𝟒 𝒎, 𝒘 = 𝝅 𝒓𝒂𝒅/𝒔, 𝝓 = 𝒓𝒂𝒅 𝟒 a) A = 4 m ∵ 𝝎 = 𝟐𝝅𝒇 𝝎 𝝅 𝒇= = = 𝟎. 𝟓 𝑯𝒛 𝟐𝝅 𝟐𝝅 𝟏 𝟏 𝑻= = =𝟐𝒔 𝒇 𝟎. 𝟓 b) 𝒅𝒙 𝝅 𝒗= = −𝟒𝝅 𝐬𝐢𝐧(𝝅 𝒕 + ) 𝒅𝒕 𝟒 𝒅𝒗 𝟐 𝝅 𝒂= = −𝟒𝝅 𝐜𝐨𝐬(𝝅 𝒕 + ) 𝒅𝒕 𝟒 𝑨𝒕 𝒕 = 𝟏 𝒔 𝝅 𝒗 = −𝟒𝝅 𝐬𝐢𝐧 𝝅 + = 𝟖. 𝟖𝟗 𝒎/𝒔 𝟒 𝟐 𝝅 𝒂 = −𝟒𝝅 𝐜𝐨𝐬 𝝅 + = 𝟐𝟕. 𝟗𝟐 𝒎/𝒔𝟐 𝟒 C) 𝒗𝒎𝒂𝒙 = ± 𝑨𝒘 = ± 𝟒𝛑 = ±𝟏𝟐. 𝟓𝟕 𝒎/𝒔 d) 𝒂𝒎𝒂𝒙 = ± 𝑨𝒘𝟐 = ± 𝟒𝝅𝟐 = ±𝟑𝟗. 𝟒𝟖 𝒎/𝒔𝟐 Angular velocity (𝒘) of Mass Spring System consider a block of mass m attached to the end of a spring, with the block free to move on a frictionless, horizontal surface. when the block is displaced to a position 𝑥 , the spring exerts on the block a force that is proportional to the position and given by Hooke’s law. 𝑭𝒓 = −𝒌𝒙 Where 𝒌 is the spring constant or force constant (stiffness of the spring) By applying Newton’s Second Law: 𝑭𝒓 = −𝒌𝒙 = 𝒎𝒂 𝒌 𝒂=− 𝒙 𝒎 ∵ 𝒂 = −𝒘𝟐 𝒙 𝟐 𝒌 ∴𝒘 = 𝒎 𝒌 ∴𝒘= 𝒎 𝒌 𝟐𝝅 ∴𝒘= = = 𝟐𝝅𝒇 𝒎 𝑻 Potential Energy (𝑼) of a Mass Spring System When a spring is compressed or stretched, it will try to restore its equilibrium position by exerting a force equal and opposite to the external force. The external work done in compression or stretching will store in the spring as potential energy. The potential energy can be calculated by getting the area under the applied force – displacement diagram. 𝑭 = 𝒌𝒙 𝟏 𝑼 = 𝑾𝒐𝒓𝒌 = 𝑭𝒙 𝟐 𝟏 𝑼 = 𝒌𝒙 𝒙 𝟐 𝟏 𝟐 𝑼 = 𝒌𝒙 𝟐 Total Mechanical Energy (𝑬) of Mass Spring System The total Mechanical Energy (𝑬) of a Mass Spring System is the sum of the Kinetic energy (𝑲𝑬) of the block and the potential energy (𝑼) stored in the spring. 𝟏 𝟐 𝟏 𝟐 𝑼 = 𝒌𝒙 = 𝒌𝑨 𝒄𝒐𝒔𝟐 𝒘𝒕 + 𝝓 𝟐 𝟐 𝟏 𝟏 𝑲𝑬 = 𝒎𝒗 = 𝒎𝒘𝟐 𝑨𝟐 𝒔𝒊𝒏𝟐 𝒘𝒕 + 𝝓 𝟐 𝟐 𝟐 𝒌 𝟐 𝒌 ∵𝒘= ⟹ 𝒘 = ⟹ 𝒌 = 𝒎𝒘𝟐 𝒎 𝒎 𝟏 𝟐 ∴ 𝑲𝑬 = 𝒌𝑨 𝒔𝒊𝒏𝟐 𝒘𝒕 + 𝝓 𝟐 𝑬 = 𝑼 + 𝑲𝑬 𝟏 𝟏 𝑬= 𝒌𝑨𝟐 𝒄𝒐𝒔𝟐 𝒘𝒕 + 𝝓 + 𝒌𝑨𝟐 𝒔𝒊𝒏𝟐 𝒘𝒕 + 𝝓 𝟐 𝟐 𝟏 𝑬= 𝒌𝑨𝟐 [𝒄𝒐𝒔𝟐 𝒘𝒕 + 𝝓 + 𝒔𝒊𝒏𝟐 𝒘𝒕 + 𝝓 ] 𝟐 ∵ 𝒄𝒐𝒔𝟐 𝒘𝒕 + 𝝓 + 𝒔𝒊𝒏𝟐 𝒘𝒕 + 𝝓 = 𝟏 𝟏 𝟐 ∴ 𝑬 = 𝒌𝑨 𝟐 The velocity of the block (𝒗) at any position (𝒙) can be determined as 𝑬 = 𝑼 + 𝑲𝑬 𝟏 𝟐 𝟏 𝟐 𝟏 𝒌𝑨 = 𝒌𝒙 + 𝒎𝒗𝟐 𝟐 𝟐 𝟐 𝒌𝑨𝟐 = 𝒌𝒙𝟐 + 𝒎𝒗𝟐 𝒎𝒗𝟐 = 𝒌𝑨𝟐 − 𝒌𝒙𝟐 𝟐 𝒌 𝟐 𝒌 𝟐 𝒗 = 𝑨 − 𝒙𝟐 ⟹𝒗=± 𝑨 − 𝒙𝟐 𝒎 𝒎 𝒌 𝒗=± (𝑨𝟐 − 𝒙𝟐 ) = ±𝒘 (𝑨𝟐 − 𝒙𝟐 ) 𝒎 Example An oscillator consists of a block attached to a spring (𝒌 = 𝟒𝟎𝟎 𝑵/𝒎). At some time 𝒕, the position (measured from the system's equilibrium location), velocity and acceleration of the block are 𝒙 = 𝟎. 𝟏 𝒎, 𝒗 = −𝟏𝟑. 𝟔 𝒎/𝒔 , 𝐚𝐧𝐝 𝒂 = −𝟏𝟐𝟑 𝒎/𝒔𝟐. Calculate (a) the frequency of motion, (b) the mass of the block, (c) the amplitude and (d) the total energy of the oscillator. Solution: a) ∵ 𝒂 = −𝒘𝟐 𝒙 ∴ −𝟏𝟐𝟑 = −𝟎. 𝟏 𝒘𝟐 ∴ 𝒘𝟐 = 𝟏𝟐𝟑𝟎 ∴ 𝒘 = 𝟏𝟐𝟑𝟎 = 𝟑𝟓. 𝟎𝟕 𝒓𝒂𝒅/𝒔 ∵ 𝝎 = 𝟐𝝅𝒇 𝝎 𝟑𝟓. 𝟎𝟕 𝒇= = = 𝟓. 𝟓𝟖 𝑯𝒛 𝟐𝝅 𝟐𝝅 b) 𝒌 𝒌 𝒌 𝟒𝟎𝟎 𝒘= ⟹ 𝒘𝟐 = ⟹𝒎= 𝟐= 𝟐 = 𝟎. 𝟑𝟑 𝑲𝒈 𝒎 𝒎 𝒘 𝟑𝟓. 𝟎𝟕 c) ∵ 𝑬 = 𝑼 + 𝑲𝑬 𝟏 𝟐 𝟏 𝟐 𝟏 ∴ 𝒌𝑨 = 𝒌𝒙 + 𝒎𝒗𝟐 𝟐 𝟐 𝟐 𝟐 𝟐 𝒌𝒙 + 𝒎𝒗 ∵ 𝑨𝟐 = 𝒌 𝒌𝒙𝟐 + 𝒎𝒗𝟐 (𝟒𝟎𝟎 × 𝟎. 𝟏 𝟐 ) + (𝟎. 𝟑𝟑 × −𝟏𝟑. 𝟔 𝟐 ) 𝑨= = = 𝟎. 𝟒 𝒎 𝒌 𝟒𝟎𝟎 𝟏 𝟏 d) 𝑬 = 𝒌𝑨𝟐 = × 𝟒𝟎𝟎 × (𝟎. 𝟒)𝟐 = 𝟑𝟐 𝑱 𝟐 𝟐 Simple pendulum The simple pendulum consists of a bob of mass (𝑚) connected to a light string of length (𝐿). When the mass (𝑚) is displaced sideways from its equilibrium position, it subjects to a restoring force (𝐹𝑟 ) due to gravity. 𝑭𝒓 = −𝒎𝒈 𝐬𝐢𝐧 𝜽 Where: 𝑔 is the acceleration due to gravity (𝑔 = 9.8 𝑚/𝑠 2 ). For small amplitudes, where θ is small, we can use the small angle approximation, in which sin 𝜃 ≈ 𝜃. Therefore, the restoring force becomes 𝑭𝒓 = −𝒎𝒈𝜽 𝒙 𝐓𝐡𝐞 𝐚𝐫𝐜 𝐥𝐞𝐧𝐠𝐭𝐡 𝒙 = 𝑳𝜽 ⟹ 𝜽 = 𝑳 𝒙 𝑭𝒓 = −𝒎𝒈 = 𝒎𝒂 𝑳 𝒈 𝒂=− 𝒙 𝑳 ∵ 𝒂 = −𝒘𝟐 𝒙 𝒈 𝒈 ∴ 𝒘𝟐 = ⟹ 𝒘= 𝑳 𝑳 𝒈 𝟐𝝅 𝒘= = = 𝟐𝝅𝒇 𝑳 𝑻 Example: Find the periodic time and frequency of a simple pendulum 1 m long. Where 𝒈 = 𝟗. 𝟖 𝒎/𝒔𝟐. Solution: 𝒈𝟗. 𝟖 𝒘= = = 𝟑. 𝟏𝟑 𝒓𝒂𝒅/𝒔 𝑳 𝟏 𝟐𝝅 𝟐𝝅 𝟐𝝅 𝒘= ⟹ 𝑻= = = 𝟐. 𝟎𝟎𝟕 𝒔 𝑻 𝒘 𝟑. 𝟏𝟑 𝟏 𝟏 𝒇= = = 𝟎. 𝟒𝟗𝟖 𝑯𝒛 𝑻 𝟐. 𝟎𝟏 Example A mass – spring system with a spring constant 1200 N/m is mounted on a horizontal table as shown in the figure. A mass of 3 kg is attached to the free end of the spring. The mass is then pulled sideways a distance of 2 cm and released. Determine (i) the frequency of oscillations, (ii) the period, and (iii) the maximum acceleration of the mass. Solution 𝒌 𝟏𝟐𝟎𝟎 𝒘= = = 𝟐𝟎 𝐫𝐚𝐝/𝐬 𝒎 𝟑 𝝎 = 𝟐𝝅𝒇 𝝎 𝟐𝟎 𝒇= = = 𝟑. 𝟐 𝑯𝒛 𝟐𝝅 𝟐𝝅 𝟏 𝟏 𝑻= = = 𝟎. 𝟑𝟏 𝒔 𝒇 𝟑. 𝟐 𝒂𝒎𝒂𝒙 = ± 𝑨𝒘𝟐 = ± 𝟎. 𝟎𝟐 × (𝟐𝟎)𝟐 = ± 𝟖 𝒎/𝒔𝟐 Example A mass of 0.5 kg connected to a light spring of force constant 20 N/m oscillates on a horizontal frictionless surface. The amplitude of the motion is 3 cm. Calculate: a) The total energy of the system. b) The maximum speed of the mass. c) The kinetic and potential energies of the system when the displacement equals 2 cm. Solution 𝟏 𝟐 𝟏 𝑬= 𝒌𝑨 = × 𝟐𝟎 × (𝟎. 𝟎𝟑)𝟐 = 𝟗 × 𝟏𝟎−𝟑 𝑱 𝟐 𝟐 𝒌 𝟐𝟎 𝒘= = = 𝟔. 𝟑𝟐 𝐫𝐚𝐝/𝐬 𝒎 𝟎. 𝟓 𝒗𝒎𝒂𝒙 = ± 𝑨𝒘 = ± 𝟎. 𝟎𝟑 × 𝟔. 𝟑𝟐 = ± 𝟎. 𝟏𝟗 𝒎/𝒔 𝟏 𝟐 𝟏 𝑼 = 𝒌𝒙 = × 𝟐𝟎 × (𝟎. 𝟎𝟐)𝟐 = 𝟒 × 𝟏𝟎−𝟑 𝑱 𝟐 𝟐 𝒗 = ±𝒘 𝑨𝟐 − 𝒙𝟐 = ± 𝟔. 𝟑𝟐 (𝟎. 𝟎𝟑)𝟐 −(𝟎. 𝟎𝟐)𝟐 = ± 𝟎. 𝟏𝟒 𝒎/𝒔 𝟏 𝟏 𝑲𝑬 = 𝒎𝒗𝟐 = × 𝟎. 𝟓 × (± 𝟎. 𝟏𝟒)𝟐 = 𝟒. 𝟗 × 𝟏𝟎−𝟑 𝑱 𝟐 𝟐 Example A 200-g block connected to a light spring for which the force constant is 5.00 N/m. It is free to oscillate on a frictionless, horizontal surface. The block is displaced 5.00 cm from equilibrium and released from rest as in Figure. (A) Find the period of its motion. (B) Determine the maximum speed of the block. (C) What is the maximum acceleration of the block? (D) Express the position, velocity, and acceleration as functions of time in SI units. Solution: Example A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body suspended from this balance, when displaced and released, oscillates with a period of 0.6 sec. what is the weight of the body? Solution Assignment A 0.15 - kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the object is 0.012 m from its equilibrium position, it is observed to have a speed of 0.3 m/s. What are (a) the total energy of the object at any point of its motion; (b) the amplitude of the motion; (c) the maximum speed attained by the object during its motion?

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