Introduction to Class XII Physics - 2024 PDF

MOMO/DUMPLING PHYSICS INTRODUCTION TO CLASS XII PHYSICS - 2024 Oscillation (12%) 864 mins 2024 Scope 1. Periodic and Oscillatory Motions. Periodic and oscillatory motions Simple harmonic motion Time period, frequency, and displacement of periodic motion and its calculation Relation between simple harmonic motion and uniform circular motion 2. Velocity and acceleration in simple harmonic motion Velocity and acceleration in simple harmonic motion Objectives Explain periodic motion, oscillatory motion and motion of simple harmonic systems graphically using interactive simulation. Periodic motion: is a motion which repeats after definite time interval. Periodic motion can be along any path. Example: pendulum bob, ball rolling back and forth in a round bowl, motion of hands of clock Oscillatory motion: is a motion in which a body moves back and forth repeatedly about a fixed point in a definite time interval. Example: pendulum bob If a periodic motion follows the same path through a fixed point, then the motion is oscillatory. Therefore, periodic motion may or may not be oscillatory but oscillatory motion is always periodic. Discussion time! (2 mins) “A bird flapping its wings circles around a clock tower". Which part of the motion is periodic and oscillatory? A. both the bird’s motion and wings flapping are non-oscillatory B. both the bird's motion and wings flapping are oscillatory C. bird’s motion is oscillatory while wings flapping is periodic D. bird’s motion is periodic while wings flapping is oscillatory D. bird’s motion is periodic while wings flapping is oscillatory Characteristic of oscillatory motion: 1. It has tendency to return to equilibrium or mean position after being disturbed. 2. Restoring force is always proportional and opposite to the displacement. 3. Energy is conserved in this motion. Simple harmonic motion: Simple harmonic motion is the motion executed by a particle subjected to a force that is proportional to the displacement of the particle but opposite in sign. Discussion time! (2 mins) i. Which of the following examples represent periodic but not simple harmonic motion? (a) the rotation of earth about its axis. (b) motion of an oscillating mercury column in a U-tube. (c) motion of a ball bearing inside a smooth curved bowl, when released from a point slightly above the lower most point. (d) A freely suspended bar magnet displaced from its N-S direction and released. A a and d B a and c Ca Dd Ca 1. Motion of seconds hand of a watch. 2. Motion of fan blades which are rotating with constant angular velocity 'w'. periodic motion but not oscillatory. Examples for oscillatory motion which are periodic: (i) Swinging a pendulum in a clock. (ii) To and fro motion of atoms in a substance. (iii) The vibration of strings in a guitar. Examples of periodic motion which are not oscillatory: (i) Uniform circular motion (ii) The motion of planets in the solar system is orbital. Simple harmonic motion as the projection of uniform circular motion: SHM is the projection of uniform circular motion on a diameter of the circle in which the circular motion occurs. For a SHM the graph of position vs time is a wave! Time period (T): is the time taken for one complete oscillation or rotation. In one complete rotation, angular displacement, θ=2Π  2  d d T= =  v= t =     t v  is angular velocity Frequency(f): Number of oscillations completed in 1 second. SI unit is hertz or per second. 1 f =............ T 2 T=  Replace value of T in equation 1 1 f =   = 2 f ( 2  ) FREQUENCY f: Number of oscillation in one second. Unit is hertz or per second Time period T: time taken for one complete oscillation. Unit is second. 1 𝑓= 𝑇 Activity! (2 mins) An insect move its wings up and down 144 times in 3 seconds. The period of this movement is A. 0.0208 seconds C. 48 seconds B. 48 Hertz D. 144 Hertz A. 0.0208 seconds Displacement equation of SHM: This equation gives displacement of a particle in SHM at any instant of time. From the diagram, y sin  = A y = A sin  y = A sin t Amplitude(A): is the maximum displacement from the mean or equilibrium position. When a particle reaches extreme position A, it undergoes maximum displacement, so θ=900  y = A sin 90o y=A Velocity in SHM: Acceleration in SHM: v =  A cos t...... y = A sin t......... Differentiating 1 with respect to time t Differentiating 1 with respect to time dv d dy d = A sin t =  A cos t dt dt dt dt d d v = A sin t a =  A cos t dt dt v = A cos t   a =  A(− sin t )   v =  A cos t a = − 2 A sin t V max = ωA a max = ω2A Discussion time! (2 mins) A particle is undergoing a SHM of amplitude 10cm. What should be the maximum value of acceleration at an extreme position for maximum speed at centre to be 5m/s? A. 20m/s2 B. 5m/s2 C. 0 m/s2 D. 250 m/s2. Answer: D Graphical representation of displacement, velocity and acceleration for SHM in sine (when the pendulum bob is at the mean position initially or when the projection is on vertical diameter) 𝑣 = 𝜔 𝐴2 − 𝑥 2 Graphical representation of displacement, velocity and acceleration for SHM in cosine (when the pendulum bob is at the extreme position initially or when the projection is on horizontal diameter) Time 0 T/4 T/2 3T/4 T Displacement A 0 -A 0 A X=A cos ωt Velocity 0 -ωA 0 ωA 0 V=-ωAsin ωt Acceleration - ω2A 0 ω2A 0 -ω2A a=-ω2Acosωt Displacement, velocity and acceleration at mean and extreme positions in SHM Phase: Phase expresses the position and direction of motion of the particle at that instant. The quantity φ is called the phase constant. It is determined by the initial conditions of the motion. If at t = 0 the object has its maximum displacement in the positive x-direction, then φ = 0, if it has its maximum displacement in the negative x-direction, then φ = π. If at t = 0 the particle is moving through its equilibrium position with maximum velocity in the negative x-direction then φ = π/2. The quantity ωt + φ is called the phase. X=A sin 𝜔𝑡 V=𝜔𝐴𝑐𝑜𝑠𝜔𝑡 1. A particle executing SHM has a maximum 𝑠𝑖𝑛2 𝜔𝑡 + 𝑐𝑜𝑠 2 𝜔𝑡 =1 cos 𝜔𝑡= 1 − 𝑠𝑖𝑛2 𝜔𝑡 displacement of 4 cm and its acceleration at a distance of 1 cm from its mean position is 3 V=𝜔𝐴 1 − 𝑠𝑖𝑛2 𝜔𝑡 cm/s2. What will be its velocity when it is at a distance of 2 cm from the mean position? 𝑥 V=𝜔𝐴 1 − (𝐴)2 6 cm/s 𝐴2 − 𝑥 2 𝑣 = 𝜔𝐴 𝐴2 𝑣 = 𝜔 𝐴2 − 𝑥 2 Which of the following relationships are found between the acceleration ‘𝛽′ and the displacement ‘y’ of a particle involved in simple harmonic motion? A. β = 3x B. 𝛽 = -5 𝑦 C. β = -50y 2 β = 0.6y 3 A. 𝛃 = -5 𝐲 The relationship between acceleration (𝛽) and displacement (y) of a particle involved in simple harmonic motion is given by the following equation: 𝛽 = -ω^2 y where 𝛽 is the acceleration of the particle, y is the displacement of the particle from its equilibrium position, and ω is the angular frequency of the motion. An object attached to a spring vibrates with simple harmonic motion as described by above Figure. For this motion, find (a) the amplitude, (b) the period, (c) the angular frequency, (d) the maximum speed, (e) the maximum acceleration, and (f) an equation for its position x as a function of time. Find the amplitude, frequency and time period of an object oscillating at the end of a spring, if the equilibrium for its position at any instant t is given by x=0.25 cos〖π/8 t〗. Find the displacement of the object after 2.0 seconds. The displacement equation of a particle executing simple harmonic motion is 𝑦=0.2𝑠𝑖𝑛 (50𝜋𝑡 + 1.57). Find the i) amplitude ii)angular frequency iii) frequency and iv) the time period. The displacement – time data of a body undergoing simple harmonic motion is given below. What is its amplitude acceleration? A. 10 cm/s2 B. 6 cm/s2 C. 24 cm/s2 D. 0.7 cm/s2 Linear combination of sine and cosine function: replace 2 and 3 in 1 z (t ) = D cos  sin t + D sin  cos t z (t ) = D sin(t +  )......this is the resultant displacement phase difference or phase angle is B B let x(t ) = A sin t and y (t ) = B cos t tan =   = tan −1   when these 2 waves are superimposed, we get A  A z(t) =x(t ) + y (t ) Resultant amplitude is z (t ) = A sin t + B cos t.......... D= A 2 + B 2 where D is the resultant amplitude The time period of the combination of the 2 or more from the triangle, periodic functions is equal to the minimum time period A cos =  D cos  = A......... among the periodic functions which have been combined D B sin  =  D sin  = B............ D i. The displacement of a particle is given by: 𝑥 = 6𝑐𝑜𝑠𝜔𝑡 + 8𝑠𝑖𝑛𝜔𝑡 in meter. Calculate the amplitude of the resultant SHM and phase angle. Energy in SHM: This restoring force is associated with potential energy (PE) and PE depends on how much the spring is stretched or compressed. Therefore, Potential energy is given by According to Hooke’s law, restoring force is 1 2 PE = kx..... always proportional to displacement and 2 opposite in direction. x = A cos t By Hooke's law, Replace in Kinetic energy (KE) F=-kx...... where k is the spring constant 1 PE = k ( A cos t ) 2 1 2 By Newton's law, KE = mv... 2 2 F=ma..... v = − A sin t.......... = 1 -kx=ma PE = kA2 cos 2 t Replace 2 in 1 2 1 -k= ma KE = m(− A sin t ) 2 x 2 we know that a=- 2 A at extreme position(x=A) 1 KE = m 2 A2 sin 2 t 2 m(- 2 A) 1 KE = kA2 sin 2 t ( m 2 = k )  -k=  k = m 2 2 A a) What type of spring would you use to increase car’s stability? Explain your answer. Less stiff (more flexible/soft/delicate) because for more stiff spring, value of k will be greater which means that restoring force will be high as per the Hooke’s law, F= − kx. So, instead of car being closer to the ground, it will be displaced higher from the ground due to restoring force. For greater stability should the spring of the car have greater or smaller spring constant For greater stability, the spring of the car should have a greater spring constant. The spring constant is a measure of how stiff the spring is, and it determines how much force is required to compress or extend the spring by a certain distance. A higher spring constant means that more force is required to compress or extend the spring, and this translates to a stiffer suspension system. A stiffer suspension system will resist the car's weight transfer during cornering and will reduce body roll, resulting in greater stability. A softer suspension system with a lower spring constant will allow more body roll and weight transfer, which can compromise stability and cause the car to feel less responsive to steering inputs. It's worth noting that while a stiffer suspension can improve stability, it can also make the car feel harsher and less comfortable to drive. It's important to strike a balance between stiffness and comfort when setting up a car's suspension system. Additionally, the spring constant should be matched to the weight of the car and other factors such as the tire characteristics, driving conditions, and the driver's preferences. Total energy (TE) Energy in case of simple pendulum TE = PE + KE 1 2 1 TE = kA cos 2 t + kA2 sin 2 t 2 2 1 2 TE = kA (cos 2 t + sin 2 t ) 2 1 TE = kA2 2 This shows that KE is maximum and PE is minimum at mean position. At extreme position, PE is maximum and KE is minimum. i) A body of mass 0.1 Kg is executing S.H.M according to the equation 3𝜋 𝑥 = 0.5cos(100𝑡 + ) meter. Find 4 a) Frequency of oscillation b) Maximum acceleration c) Total energy i. When the potential energy and kinetic energy are equal, the amplitude ‘A’ of motion of a 𝐴 particle in SHM is ±. T/F 2 The laboratory worksheet recorded timing of 20 oscillations of the spring instead of just one oscillation. This is because the period of oscillation is expected to vary. T/F A student designs a toy that undergoes simple harmonic motion with amplitude A. She records the variation of mechanical energy of the toy at different positions and obtains a graph as given in the figure. At what distance from the mean position does its kinetic energy become equal to its potential energy? [BHSEC, 2021] A 2 A 2 2A 2A = Graphical representation of energy in SHM. x TIME PERIOD OF A PENDULUM IN SHM: ma = mg L x Restoring force, a=g......... L F = mg sin  displacement(x) Assuming  to be very small, we have We know that T=2 or acceleration(a) F = mg......... x From the diagram, T = 2.......... a x  =... Replace in L x Replace (2) in (1) T = 2 x x g F = mg......... L L L By Newton's law, F=ma....... T = 2 g This shows that T depends on length (L) but not on mass (m) of the bob. A hollow sphere is filled with water through a small hole at the bottom. It is hung by a long thread and as water slowly flows out of the hole, the period of oscillations first increases and then decreases. Explain why? When the sphere is filled with water, its C.G. is at the center of the sphere. We know that the time period (T) of oscillation is directly proportional to the square root of the effective length of the pendulum. i.e. T ∝ √l.---------------------------------------------------------------------------------------------------------------------------------------[0.5] Initially, the centre of mass of the sphere is at the centre of the sphere. As the water slowly flows out of the hole at the bottom, the CM of the liquid (hollow sphere) first goes on downward and then upward. Hence, the effective length of the pendulum first increases and then decreases.----------------------------------------------------------------------------------------------------------------[1.5] OR Only formula-----0.5 OR No formula but clear explanation----------2 The figure below shows the favorite game of Miss Dema. Use the below figure to answer the questions that follows; What would happen to the time period of a swing if Miss Dema swinging in sitting position stands? Why? What would have happened to the time period if she swings in the same swing at higher altitude than before? Why? Miss Dema’s close friend came to play with her, what would be the effect in the time period of a swing if Miss Dema allowed her friend? Time period of a body undergoing Horizontal SHM: By Hooke's law, F = −kx........ By Newton's 2nd law, F = ma.... = ma = kx kx a =.......... m x We know that T = 2....... a Replace in x T = 2 kx m m T = 2 k A particle of mass 2 kg moves in simple harmonic motion and its potential energy U varies with position x as shown. The period of oscillation of the particle is: The time period of a simple pendulum is given by https://wheelofnames.com/ Where, l is the length of the pendulum from the point from suspension and g is the acceleration due to gravity. When the pendulum is taken to the Moon, the value of ‘g’ decreases and the time period increases. Thus, the pendulum takes more time to complete one vibration. Since the time period increases the speed of oscillation of the pendulum also decreases. 10.A force of 0.4 N is required to displace a body attached to a spring through 0.1 m from its mean position. Calculate the spring constant of the spring. 12. A body of mass 0.025 kg attached to a spring is displaced through 0.1 m to right of the mean position, if spring constant is of the spring is 0.4 N/m and its velocity at the end of this displacement be 0.4 m/s. Calculate (i) Time period (ii) Frequency (iii) Angular speed w (iv) Total energy (v) The amplitude (vi) Maximum velocity (vii) The maximum acceleration A 326-g object is attached to a spring and executes simple harmonic motion with a period of 0.250 s. If the total energy of the system is 5.83 J, find (a) the maximum speed of the object, (b) the force constant of the spring, and (c) the amplitude of the motion. A block weighing 4 kg extends a spring by 0.16 m from its un- stretched position. The block is removed and a 0.50 kg body is hung from same spring. If the spring is now stretched and released, what is its period of vibration? A simple pendulum is made of a body which a hollow sphere is containing mercury suspended by means of a wire. If a little mercury is drained off, the period of pendulum will A) Remains unchanged B) Increase C) Decrease D) Become erratic Correct Answer: B Class XII science students did an experiment on simple pendulum in the physics laboratory using 70cm long thread and bob of 1.2cm in diameter. On setting in oscillation, the bob took 35 sec to complete 20 oscillations. (i) What will be the acceleration due to gravity in the physics laboratory? (ii) What is the difference between the standard value and your value? [1⁄2] 3.92 cm Free oscillation: Free oscillation is the oscillation of the body if it vibrates with its own natural frequency without the help of any external periodic force. Example: A person swinging in a swing without anyone pushing. Forced or driven oscillation: Forced or driven oscillation is the oscillation of a body if it vibrates with the help of external periodic force with frequency different from the natural frequency of the body. Example: a person pushing the swing periodically. There are 2 angular frequencies associated with a system undergoing forced oscillation: 1. Natural angular frequency (ω): this is the frequency at which a body would oscillate if it was suddenly disturbed and then left to oscillate freely. 2. Angular frequency (ωd): this is the frequency of the external driving force which causes the forced oscillation. NATURAL FREQUENCY The natural frequency of a body is a characteristic frequency at which a system or object vibrates when it is disturbed from its equilibrium state. In simpler terms, it is the frequency at which an object naturally oscillates or vibrates without any external force or disturbance. The natural frequency of a body depends on various factors such as its size, shape, and material properties. For example, a pendulum has a natural frequency of oscillation that depends on its length and gravitational acceleration. Similarly, a guitar string has a natural frequency of vibration that depends on its tension, length, and mass per unit length. Understanding the natural frequency of a body is important in various fields of engineering and science, including mechanical and civil engineering, physics, and acoustics, as it can help in designing systems that are more efficient and less prone to damage due to vibrations. Resonance: The phenomenon of increase in the amplitude of oscillation when the frequency of the driving force is very close to the natural frequency of the body or oscillator is called resonance. For maximum amplitude displacement: 1. ω = ωd 2. damping constant (b, which depends on the fluid in which the body is oscillating)should be minimum. CBQ 1. Why do aircraft designers make sure that the natural frequency of the wings do not match with the angular frequency of the aircraft engine? Ans: because if the natural frequency of the wing matches with the angular frequency of the engine then the wings will flap violently which would be obviously dangerous. The earthquake with a magnitude of 6.1 on the Richter scale occurred in 2009 in Bhutan destroying many houses in the Eastern part of the country. Which of the following concepts can best explain the movement of tectonic plates? A. resonance B. periodic Motion C. oscillatory Motion D simple Harmonic Motion A. resonance Why are soldiers in general ordered to “route step” (walk out of step) across a bridge? The bob of a vibrating pendulum is made of ice. How will the time period change when the ice starts melting? i. What would happen to time period of a swing if a girl swinging in sitting position stands? Why? [1.5] ii. What would happen to the time period if she swings in the same swing at higher altitude than before? Why? [1.5] 1.The displacement vs time graphs of 2 SHMs are given below. Which parameter is the same for both of them? 1.A particle is performing a SHM. Calculate its total energy if its mass is doubled keeping the amplitude and force constant the same.

Introduction to Class XII Physics - 2024 PDF

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