Lesson One Oscillatory Motion Physics 2 Mr. Ahmed El Bagoury – A branch of science which concerned mainly with matter in relation to energy. – Physics concerns with physical and universal phenomena and studying these phenomena using rules. – The rules consist of physical quantities. 3 Physical Quantities Scalar Quantity Vector Quantity Mr. Ahmed El Bagoury Scalar Quantity 4 – t is a physical quantity that is defined by its magnitude only. – Such as: Mass. Length. Time. Energy. Mr. Ahmed El Bagoury Vector Quantity 5 – It is a physical quantity that is defined by its magnitude and direction. – Such as Displacement. Force. Velocity. Mr. Ahmed El Bagoury Kinds of motion Transolatory Motion • Motion from one position to another point Periodic Motion • motion of an object that regularly returns to a given position after fixed time interval 7 Hook’s law – The extension is directly proportional to the stretching force 𝑭𝒐𝒓𝒄𝒆 𝜶 𝑬𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝑭𝒐𝒓𝒄𝒆 = 𝑺𝒑𝒓𝒊𝒏𝒈 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 ×𝑬𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝑭 = 𝑲 .𝑿 𝑭 𝑲= 𝑿 𝑺𝑰 𝒖𝒏𝒊𝒕: 𝑵/𝒎 A. Bagoury 8 Spring Constant 𝑭 𝑲= 𝑿 – it is the force per unit extension. – it shows how the spring resists being extended. – It measures the stiffness of the spring – The spring constant varies from one spring to another due to the difference in:  Type of the material. Thickness of spring. 𝑮𝒓𝒂𝒅𝒊𝒆𝒏𝒕 = 𝑺𝒑𝒓𝒊𝒏𝒈 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 ∆𝑭 = ∆𝑿 A. Bagoury  Length of the spring. Thickness of wire from which the spring is made.  Number of turns of the spring Series Springs: – Each spring is loaded with the load on the system. – The total extension of the system is the sum of each individual extension of each spring – For each spring 𝑿𝑺𝒚𝒔𝒕𝒆𝒎 = 𝒙𝟏 + 𝒙𝟐 + ⋯ 𝐹 𝐹 𝐹 = + +⋯ 𝑘HIJKLM 𝑘N 𝑘O 𝟏 𝒌𝒔𝒚𝒔𝒕𝒆𝒎 𝟏 𝟏 = + +⋯ 𝒌𝟏 𝒌𝟐 – If the system is only two spring: 𝒌𝑬𝒇𝒇 𝒌𝟏 𝒌𝟐 = 𝒌𝟏 + 𝒌𝟐 Parallel Springs – The load attached to the system is divided to each spring. – The total extension of the system is the same as the the extension of each. – For each spring: 𝐹 = 𝐹N + 𝐹O + 𝐹R + ⋯ 𝑘HIJKLM . 𝑥 = 𝑘N . 𝑥 + 𝑘O . 𝑥 + 𝑘R . 𝑥 𝑘TUU = 𝑘N + 𝑘O + 𝑘R PROBLEM A spring is hung vertically (Fig. 13.2a), and an object of mass ves m attached to the lower end is then slowly lowered a distance d to the equilibrium point (Fig. 13.2b). (a) Find the value of the spring constant if the spring is displaced by 2.0 cm and the mass is 0.55 kg. (b) If a second identical spring is attached to the object in parallel with the first spring (Fig. 13.2d), where is the new equilibrium point of the system? (c) What is the effective spring constant of the two springs acting as one? ertical Spring S TR This example is an application of Newton’s second law. The r with the force ofATEGY gravity and The cau of th a spring force is upward, balancing the downward force of gravity mg when the system is in equilibrium. (See Fig. 13.2c.) Because the suspended Figure g. 13.2a), and object an object is of in mass equilibrium, the forces on the SFobject sum to zero, and it’s pos- the spri s owered a distance to the equibalance sibled to solve for the spring constant d k. Part (b) is solved the same way, but ue of the spring if the forces balancing the force of gravity. The spring constants rium, it hasconstant two spring is 0.55 kg. (b)are If a known, second idenso the second law for equilibrium can be solved for the dis- two spri rallel with theplacement first spring of (Fig. the spring. Part (c) involves using the displacement found in part (b). of the system? (c)spring, What is the the second The elongation d is S lent law then leads to the mg effective spring constant of the two-spr acting as one? n of Newton’s second law. The ION nward force ofSOLUT gravity mg when caused by the weight mg of the attached object. a b c d – (a) 𝐹 = 𝑘. 𝑥 𝑚. 𝑔 = 𝑘. 𝑥 0.55×9.81 = 𝑘×2 𝑘 = 2.7 N/cm - (b) Since the springs in parallel 𝑘JIJKLM = 𝑘N + 𝑘O = 2.7×2 = 5.4 𝑁/𝑐𝑚 𝐹 = 𝑘JIJKLM . 𝑥 0.55×9.81 = 5.4 x X = 1 cm - (c) Since the springs in parallel 𝑘JIJKLM = 𝑘N + 𝑘O = 2.7×2 = 5.4 𝑁/𝑐𝑚 Elastic Potential Energy [Strain Energy] – It is the energy stored in an elastic material due to a force that cause an extension, compression, deformation or change in size of a body. 𝑊𝑟𝑜𝑘 = 𝑃𝐸LlmJKno 1 𝑃𝐸LlmJKno = 𝐹. ∆𝑥 2 𝐹 = 𝑘. ∆𝑥 𝟏 𝑷𝑬𝒆𝒍𝒂𝒔𝒕𝒊𝒄 = 𝒌. ∆𝒙𝟐 𝟐 – The work done in stretching or compressing a material is always equal to the area under the graph of force against extension. Simple Harmonic Motion – It is the motion of a oscillating object in which the acceleration (force) is directly proportional to the displacement and opposite to it. 𝑎 ∝ −𝑥 Where a is the acceleration of the oscillating body. 𝒙 is the displacement covered by the object Motion of an Object Attached to a Spring – Equilibrium position of the system, which we identify as x = 0 – During the oscillation of the mass m, the spring exerts a force F called restoring force. Measuring Oscillation Period T: – It is the time taken by the vibrating body to make one complete oscillation time T= No. of oscillations Frequency f: – It is the number of oscillations per unit time. No. of oscillations f= time 1 f= T Amplitude 𝐱𝐨 – It is the maximum displacement away from the equilibrium position. Phase – The point that an oscillating mass has reached within the complete cycle of an oscillation at certain instant with certain speed. – It is often important to describe the phase difference between two oscillations. 630 𝑝ℎ𝑎𝑠𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 × 𝜆 Restoring Force – It is the force that tends to bring the system back to the equilibrium position. 𝐹J = −𝑘. 𝑥 From Newton’s 2nd Law 𝑭𝒙 = 𝒎. 𝒂𝒙 Because the oscillating mass is vibrating the x dimension only, so the resultant force acting on it only in dimension x. 𝑚. 𝑎• − 𝑘. 𝑥 𝒌 𝒂𝒙 = − . 𝒙 𝒎 𝒌 𝒂𝒙 = − . 𝒙 𝒎 When x= +A When x= 0 When x= -A 𝒌 𝑎 = − .𝒙 𝒎 Speed=0 𝑎=𝟎 Speed=max 𝒌 𝑎 = + .𝒙 𝒎 Speed=0 Particle in Simple Harmonic Motion 𝑑𝑥 ∵𝑣= 𝑑𝑡 𝑑O𝑥 𝑎= O 𝑑𝑡 𝑑𝑣 𝑎= 𝑑𝑡 𝒌 𝒂𝒙 = − . 𝒙 𝒎 𝑑O𝑥 𝒌 = − .𝒙 O 𝑑𝑡 𝒎 The constant 𝒌 is 𝒎 represented by 𝝎𝟐 to preserve the negative sign 𝝎𝟐 𝒌 = 𝒎 𝒅𝟐 𝒙 𝟐. 𝒙 = −𝝎 𝒅𝒕𝟐 𝒈𝒓𝒂𝒅𝒊𝒆𝒏𝒕 = −𝝎𝟐 a/ m.s-2 +𝑥š −𝑥š x/m Mathematical Solution of 𝒅𝟐 𝒙 𝒅𝒕𝟐 – Function 𝒙(𝒕) that satisfies this second-order differential equation. – The trigonometry function sine, and cosine exhibit this behavior. 𝒙 𝒕 = 𝑨 𝐜𝐨𝐬 𝝎𝒕 + ∅ Both ∅, 𝒂𝒏𝒅 𝑨 are constants. = 𝟐 −𝝎 . 𝒙 Explicitly of This Solution 𝑥 𝑡 = 𝐴 cos 𝜔𝑡 + ∅ 𝑑𝑥 ∵𝑣= = −𝐴𝜔 𝑠𝑖𝑛 𝜔𝑡 + ∅ 𝑑𝑡 𝑑O𝑥 𝑎 = O = −𝐴𝜔O cos 𝜔𝑡 + ∅ 𝑑𝑡 𝒅𝟐 𝒙 𝟐. 𝒙 = −𝝎 𝒅𝒕𝟐 Graphical Representation 𝑑O𝑥 O cos 𝜔𝑡 + ∅ = −𝐴𝜔 𝑑𝑡 O 𝒅𝟐 𝒙 𝟐. 𝒙 = −𝝎 𝒅𝒕𝟐 – A, called the amplitude of the motion, is simply the maximum value of the position of the particle in either the positive or negative x direction. – The constant 𝜔 is called the angular frequency, – The constant angle ∅ is called the phase constant (initial phase angle) – phase constant is determined uniquely by the position and velocity of the particle at t =0. – The quantity 𝜔𝑡 + ∅ is called phase of the motion. Angular frequency 𝝎 – It is a measure of how rapidly the oscillations are occurring. – The more oscillations per unit time, the higher the value of 𝜔. – It has units of radians per second. 𝝎𝟐 𝒌 = 𝒎 𝝎= 𝒌 𝒎 𝝎 = 𝟐𝝅𝒇 Q2 Answer (D) 𝒇= 𝟏 𝟏 = 𝑻 𝟖 𝟐𝝅 𝝎= 𝟖 𝑥 = 𝐴 𝑠𝑖𝑛 𝜔𝑡 𝟐𝝅 4 3 𝑥 = 1 × sin × = 𝟖 3 2 © 𝑫𝒐𝒏 𝒕 𝑭𝒐𝒓𝒈𝒆𝒕 … … … … 𝝅 = 𝟏𝟖𝟎 𝟐𝝅 O 𝑎 = −𝝎 𝑥 = − 𝟖 𝜋O 3 𝑎=− 32 O 3 × 2 Equations for velocity – If we take time t = 0 when the oscillator passes through the middle of the oscillation with its greatest speed v0, then we can represent the changing 𝑣 = 𝑥š . 𝜔 𝑐𝑜𝑠 𝜔. 𝑡 (1) 𝑡=0 𝒗𝒐 = 𝒙𝒐 . 𝝎 x = x- sin ω. t x = sin ω. t x𝐱 𝐱𝐨 𝟐 = 𝐬𝐢𝐧𝟐 𝛚. 𝐭 (𝟐) ∵ cos O ω. t + sinO ω. t = 1 ∴ cos O ω. t = 1 − sinO ω. t 𝐜𝐨𝐬 𝛚. 𝐭 = 𝟏 − 𝐬𝐢𝐧𝟐 𝛚. 𝐭 (𝟑) From 2, and 3 𝐜𝐨𝐬 𝛚. 𝐭 = 𝐱 𝟏− 𝐱𝐨 𝟐 𝟒 From 1, and 4 x ∵ v = x- . ω 1 − x- O 𝒗 = 𝝎 𝒙𝒐 𝟐 − 𝒙𝟐 Kinetic Energy It is the energy associated to the moving objects. 𝟏 𝑬𝑲 = . 𝒎. 𝒗𝟐 𝟐 𝒗 = 𝝎 𝒙𝒐 𝟐 − 𝒙𝟐 𝟏 𝑬𝑲 = . 𝒎. 𝝎𝟐 . 𝒙𝒐 𝟐 − 𝒙𝟐 𝟐 at time t=0 𝑥 = 𝑥š . 𝑠𝑖𝑛 𝜔. 𝑡 𝑥=0 1 𝐸¶ = . 𝑚. 𝜔O . 𝑥š O 2 Potential Energy: – it is the work done against the restoring force at a given instant. 1 𝑤𝑜𝑟𝑘 = 𝐸¸ = . 𝐹¹LJKš¹nº» . 𝑥 2 𝑭𝒓𝒆𝒔𝒕𝒐𝒓𝒊𝒏𝒈 = −𝒎. 𝒂 = 𝒎. 𝝎𝟐 . 𝒙 1 𝐸¸ = . 𝑚. 𝜔O . 𝑥. 𝑥 2 𝟏 𝑬𝒑 = . 𝒎. 𝝎𝟐 . 𝒙𝟐 𝟐 Total Energy – It is the sum of the kinetic energy and potential energy. – The total energy is conserved according to the law of conservation of energy in absence of frictional forces. 𝐸KšKml = 𝐸¸ + 𝐸¶ 1 1 O O O 𝐸KšKml = . 𝑚. 𝜔 . 𝑥š − 𝑥 + 𝐸¸ = . 𝑚. 𝜔O . 𝑥. 𝑥 2 2 𝟏 𝑬𝒕𝒐𝒕𝒂𝒍 = . 𝒎. 𝝎𝟐 . 𝒙𝒐 𝟐 𝟐 Systems in S.H.M Mass-Spring System Pendulum Mass-Spring System – Since the spring obeys Hook’s law 𝑭 = −𝒌. ∆𝒙 – where k is the spring constant, ∆x is the displacement – Newton’s second law 𝐅 = 𝐦. 𝐚 𝐦. 𝐚 = −𝐤. ∆𝐱 𝐤 𝐚 = − . ∆𝐱 𝐦 – By comparing this equation with the general formula of the S.H.M 𝐚 = −𝛚𝟐 . ∆𝐱 𝒌 𝟐 𝝎 = 𝒎 Period and Frequency ω= k m ∵ ω = 2πf 2πf = k m 𝟏 𝒌 𝒇= 𝟐𝝅 𝒎 1 f= T 𝒎 𝑻 = 𝟐𝝅 𝒌 Simple Pendulum – Tension force in the string T: 𝐓 = −𝐦𝐠 𝒄𝒐𝒔 𝜽 – Restoring force F F = −mg sin θ – So we have to keep θ as small as we can at most 10o. – Because in this range of angles: 𝐬𝐢𝐧 𝛉 𝐝𝐞𝐠𝐫𝐞𝐞 = 𝛉 𝐢𝐧 𝐫𝐚𝐝𝐢𝐚𝐧. 𝐅 = 𝐦. 𝐚 𝐅 = −𝐦𝐠 𝐬𝐢𝐧 𝛉 𝐦. 𝐚 = −𝐦𝐠 𝐬𝐢𝐧 𝛉 𝐚 = −𝐠 𝐬𝐢𝐧 𝛉 From the figure: ∆x sin θ = L 𝐱 𝐚 = −𝐠. 𝐋 a = −ωO . ∆x 𝝎𝟐 𝒈 = 𝑳 Period and Frequency ω= g L ω = 2πf 𝒈 𝟐𝛑𝐟 = 𝑳 𝟏 𝒈 𝒇= 𝟐𝝅 𝑳 1 T= f 𝑳 𝑻 = 𝟐𝝅 𝒈 Damped Oscillations – It is the gradual decrease in the energy of the oscillating body because of frictional force. 𝟏 𝑬𝒕𝒐𝒕𝒂𝒍 = . 𝒎. 𝝎𝟐 . 𝒙𝒐 𝟐 𝟐 Car Damper Oil or other viscous fluid Coil spring Piston with holes a b Shock absorber Q1 Propagation of a Disturbance Transverse • The disturbance is in a direction perpendicular to the propagation direction of the wave. Longitudinal • The disturbance is in a direction parallel to the propagation direction of the wave. Combination of Transverse and Longitudinal Displacements – Surface-water waves are a good example. – When a water wave travels on the surface of deep water, elements of water at the surface move in nearly circular paths. – The disturbance has both transverse and longitudinal components. – The transverse displacements seen represent the variations in vertical position of the water elements. – The longitudinal displacements represent elements of water moving back and forth in a horizontal direction. The Three-Dimensional Waves – The slower transverse waves, called S waves, with “S” standing for secondary, travel through the Earth at 4 to 5 km/s near the surface. – By recording the time interval between the arrivals of these two types of waves at a seismograph, the distance from the seismograph to the point of origin of the waves can be determined. – This distance is the radius of an imaginary sphere centered on the seismograph. – The origin of the waves is located somewhere on that sphere. – The imaginary spheres from three or more monitoring stations located far apart from one another intersect at one region of the Earth, and this region is where the earthquake occurred. – That travel out from a point under the Earth’s surface at which an earthquake occurs. – Both types, transverse and longitudinal are formed. – The longitudinal waves are the faster than the transverse waves, and traveling at speeds in the range of 7 to 8 km/s near the surface. – These longitudinal waves are called P waves, with “P” standing for primary, because they travel faster than the transverse waves and arrive first at a Seismo-graph (a device used to detect waves due to earthquakes). Wave Function – Figure represents the shape and position of the pulse at time t = 0. – Mathematical function: 𝑦 𝑥, 0 = 𝑓(𝑥). – This function describes the transverse position y of the element of the string located at each value of x at time t = 0. – The speed of the pulse is v. – The pulse has traveled to the right a 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = 𝑣𝑡 at the time t. Wave Function – We assume the shape of the pulse does not change with time. – At time t, the shape of the pulse is the same as it was at time t = 0 – Consequently, an element of the string at x at this time has the same y position as an element located at x - vt had at time t = 0 𝑦 𝑥, 𝑡 = 𝑦(𝑥 − 𝑣𝑡). In general, 𝒚 𝒙, 𝒕 = 𝒇(𝒙 − 𝒗𝒕). Wave function, depends on the two variables x and t Waveform – If t is fixed (as, for example, in the case of taking a snapshot of the pulse), – the wave function y(x), sometimes called the waveform, defines a curve representing the geometric shape of the pulse at that time.