Chap 11. Vibration and Waves. The impressed force on an object is proportional to its displacement from it equilibrium position.

Transcription

1 Chap 11. Vibration and Waves Sec Simple Harmonic Motion The impressed force on an object is proportional to its displacement from it equilibrium position. F x This restoring force opposes the change F = kx In case of a spring the constant k is called as the spring constant Elastic property: It takes about twice as much force to stretch/compress a spring twice as far. That linear dependence of displacement upon stretching force is called Hooke s law. If there is no friction here mechanical energy is conserved.

2 Work done to extend/compress a spring through a distance x from its equilibrium position: W = x 0 W = x 0 F dx = Potential energy P E = 1 2 kx2 = W s x 0 ( kx)dx = 1 2 kx2 2 The time taken to complete one cycle is called as the period (T ). The number of cycles per second is called as the frequency (f ) of revolution, such that f = 1 T The maximum displacement from the equilibruim point is called the Amplitude of the motion

3 Problem D: If the spring constant is k = 80N/m. A mass of 0.5 Kg is held at 2 cm and released. Describe the motion and find the velocity at x = 0 Soln D: Energy is conserved Spring P E 1 + KE 1 = Spring P E 2 + KE kx2 i mv i 2 = 1 2 kx2 f mv f kx2 i + 0 = mv f 2 (1) 3 V f = x i k m = 0.253m/s Sec Energy in a Simple Harmonic Oscillator Consider the motion of a simple pendulum. The total mechanical energy at any point is conserved. Hence we can use this fact to obtain a relation for the velocity, amplitude etc.. for the SHM. The total energy at the highest point (amplitude(a)) is given as. T.E A = KE + P E = ka2 The total energy at the lowest point (A=0) is given as. T.E 0 = KE + P E = 1 2 mv2 max + 0 The total energy at a distance (x) with velocity (v) is given as T.E 0 = KE + P E = 1 2 mv kx2

4 Using the above the relation obtained from energy conservation we get the following: k v max = m A = ωa We can also deduce 1 2 mv kx2 = 1 2 ka2 v = ±v max 1 ( x A )2 F max = ka = ma max a max = k m A a max = ω 2 A Sec. 11.3A- Period and Frequency of a S.H.Oscillator The SHM can be considered as a motion were the object is revolving in a circle of radius A with a velocity v max. In such a motion We derived earlier that This gives v max = 2πA T v max = T = 2πA v max k m A = ωa T = 2πA = 2πA m = 2π v max k m A k = 2π ω Frequency can be obtained as f = 1 T = 1 k 2π m = ω 2π 4

5 Sec. 11.3B- Mathematical Representation of a wave 5 If a object undergoes a circular motion, then its projection along the x axis undergoes a SHM. The function x = Acosθ or x = Asinθ gives the displacement of a wave at any time t The motion will be a SHM if F x: Proof: We have by definition of angular velocity ω = θ/t x = Acos(ωt) = x max cos(ωt) v = dx dt = Aωsin(ωt) = v maxsin(ωt) a = d2 x dt = 2 Aω2 cos(ωt) = a max cos(ωt) F = ma = mω 2 x = kx This gives x = Asin(ωt) = x max sin(ωt) v = dx dt = Aωcos(ωt) = v maxcos(ωt) a = d2 x dt = 2 Aω2 sin(ωt) = a max sin(ωt) F = ma = mω 2 x = kx x max = A k v max = Aω = A m a max = Aω 2 = A k m ω = 2πf = 2π T Look at the sin wave plotted earlier

6 Which equation should one use to represent a wave?. ANS: Cosine or sine function can be used by correctly defining the phase angle. Use the initial condition given to you and that will decide the phase angle. Sec Simple Pendulum 6 A simple pendulum is one which can be considered to be a point mass m suspended from a string or rod of negligible mass. Let the length of the string be L. The restoring force in this case is F = kx = mgsinθ = mg x L since sinθ = x L This gives T = 2π k = ω 2 m = mg L L g ω = 2π T = g L Period of a simple pendulum

7 Problem A: The mass of 0.5 kg is performing a SHM represented as x = 3 sin (4t) Find all the relevant quantities. Soln A This equation is of the form x = 3 sin (4t) = Asin(ωt) Amplitude A = 3 m Angular velocity ω = 4 rad/s Maximum velocity v max = Aω = 12m/s Maximum acceleration a max = Aω 2 = 48m/s 2 Total Energy T E = 1 2 mv2 max = = 36 J The spring constant k = ω 2 m = = 8 N/m What is the velocity, PE and KE at x = 2.5 v = ±v max 1 ( x A )2 = ±12 1 ( )2 = ±6.63m/s KE = 1 2 mv2 = = 11J P E = T E KE = = 25 J Problem B: Find the period of a simple pendulum of length 0.5 m Soln B T = 2π = 1.41s 7

8 Sec Wave Motion 8 Waves carry energy and not particles. Every point on the wave performs a SHM. Hence the velocity of the wave is different from the velocity of the particle Velocity of the wave is defined as v = λ T Wavelength λ of the particle is defined as the distance between two adjacent crest or trough. Now you plot (v vs x) to see the same. Transverse Waves: Here the oscillation of the particles are perpendicular to the direction of motion of the wave. Eq. Light. They travel in terms or crest and trough Longitudinal Waves: Here the oscillation of the particles are parallel to the direction of motion of the wave. Eq. Sound. They travel in a medium in the form of compression and rarefactions. A wave in a slinky is a good visualization.

9 Sec Damped Harmonic motion READ FROM TEXT Sec Forced Vibrations: Resonance A resonant frequency is a natural frequency of vibration determined by the physical parameters of the vibrating object. This same basic idea of physically determined natural frequencies applies throughout physics in mechanics, electricity and magnetism, and even throughout the realm of modern physics. Some of the implications of resonant frequencies are: It is easy to get an object to vibrate at its resonant frequencies, hard to get it to vibrate at other frequencies. eg: Swing A vibrating object will pick out its resonant frequencies from a complex excitation and vibrate at those frequencies, essentially filtering out other frequencies present in the excitation. eg: mass attached to spring Most vibrating objects have multiple resonant frequencies. We can also apply external force and this gives rise to forced vibrations. For a spring the natural frequency of vibration is 9 f o = 1 2π k m If we apply an external force and let it vibrate with a frequency f The amplitude of the motion will be maximum when f = f o. The natural vibrating frequency f o of an object is called as the resonant frequency.

10 Sec Energy transported by waves 10 When a wave travels, matter does not move but energy is transported. This energy is proportional to the square of the amplitude. Energy = Amplitude 2 The Intensity of the wave is defined as the power transported per unit area in the direction perpendicular to the energy flow. Intensity = energy/time Area = power Area = P πr 2 Intensity is a scalar and the SI units is W att/m 2. Since Intensity(I) 1 r 2 I 1 I 2 = r2 2 r 2 1 When the distance doubles the intensity is 1/4 th its original value. Intensity(I) A I 1 I 2 = A2 1 A 2 2 = r2 2 r 2 1 Problem A: The intensity of a tremor travelling through the earth at a distance 50 km from the source is 10 4 W/m 2. What is its intensity at a distance 100 km Soln A I 1 I 2 = r2 2 r 2 1 I 1 = = W/m 2

11 Sec Reflection and Transmission of Waves 11 A wave travelling a medium when it hits a boundary of another medium can get reflected/ transmitte or both. The nature of the reflection depends on the boundary surface. The frequency of the wave does not change as the wave gets transmitted but the velocity and hence the wave length of the wave changes. A transverse wave is assumed to be travelling in the form of a wave/ wave front. Eq Ocean Wave. Hence,wavefront from far away sources are parallel to each other. A ray is a line drawn perpendicular to the wave front and gives direction of the travelling wave. Laws of Reflection The incident wave front, normal and reflected wave front all lie in the same plane. The angle of incidence is equal to the angle of reflection. i = r

12 Laws of Refraction 12 Refraction is the bending of a wave when it enters a medium where it s speed is different. The refraction of light when it passes from a fast medium to a slow medium bends the light ray toward the normal to the boundary between the two media. The amount of bending depends on the indices of refraction of the two media and is described quantitatively by Snell s Law. The incident wave front, normal and reflected wave front all lie in the same plane. The refractive index of the medium is given by Snell s Law µ = v 2 v 1 = sinθ 2 sinθ 1 As the speed of light is reduced in the slower medium, the wavelength is shortened proportionately. The frequency is unchanged; it is a characteristic of the source of the light and unaffected by medium changes.

13 Sec Interference Two traveling waves which exist in the same medium will interfere with each other. If their amplitudes add, the interference is said to be constructive interference, and destructive interference if they are out of phase and subtract. Patterns of destructive and constructive interference may lead to dead spots and live spots in auditorium acoustics. 13 Phase describes the relative positions of the interfering waves. Two waves that are in phase(crest on crest) will interfere constructively. Two waves that are out of phase (crest on trough) will interfere destructively

14 Sec Standing Waves 14 The modes of vibration associated with resonance in extended objects like strings and air columns have characteristic patterns called standing waves. Interference of incident and reflected waves is essential to the production of resonant standing waves. Interference has far reaching consequences in sound because of the production of beats between two frequencies which interfere with each other. The behavior of the waves at the points of minimum and maximum vibrations (nodes and antinodes) contributes to the constructive interference which forms the resonant standing waves. Standing waves in air columns also form nodes and antinodes, but the phase changes involved must be separately examined for the case of air columns. The medium appears to vibrate in segments or regions and the fact that these vibrations are made up of traveling waves is not apparent - hence the term standing wave.

15 Standing Waves in a String 15

16 Harmonics in a String fixed at both ends 16 Fundamental Frequency or First Harmonic: Contains two nodes at the ends and one antinode. L = λ 2 f 1 = v 2L First Overtone or Second Harmonic: Contains 3 nodes at the ends and 2 antinode. L = 2λ 2 f 2 = 2v 2L Second Overtone or third Harmonic: Contains 4 nodes at the ends and 3 antinode. In general L = 3λ 2 L = nλ 2 f 3 = 3v 2L f n = nv 2L

17 Problem B: A guitar string is 1.0 m and has a mass of 8 g. How much tension must the string be under to be able to vibrate at a fundamental frequency of f = 120 Hz. Determine the frequency of the first three overtones. Soln B The velcity can be obtained from f 1 = v 120 = v v = 240m/s 2L 2 1 The relation between the velocity and the tension (F T ) in the string is F T = m L v2 F T = = N 1 The frequency of the next three overtones are 1 st Overtone = 240 Hz, 2 nd Overtone = 360 Hz, 3 rd Overtone = 480 Hz 17 Sec Diffraction Diffraction: the bending of waves around small obstacles and the spreading out of waves beyond small openings The fact that you can hear sounds around corners and around barriers involves both diffraction and reflection of sound. Good soundproofing requires that a room be well sealed, because any openings will allow sound from the outside to spread out in the room - it is surprising how much sound can get in through a small opening. A fundamental principle of imaging is that you cannot see an object which is smaller than the wavelength of the wave with which you view it. You cannot see a virus with a light microscope because the virus is smaller than the wavelength of visible light Amount of diffraction of a light of wavelength λ through an obstacle of width L θ = λ L