FAN wxves © RIPHAH INTERNATIONAL UNIVERSITY, > Awave is any disturbance that transmits energy through matter or space. > A medium is a substance through which the wave can work. > A mechanical wave is a disturbance that travels through some material or substance called the medium for the wave. > Mechanical waves are waves that spread through a solid, liquid or gas material. The speed of the wave depends on what the medium is made of. > A mechanical wave is a wave that is a vibration in matter, transferring energy through a material.WSS Perpendicular ES ey tite arl | es ¢ Parallel a, Transverse wave > In a transverse wave the particle displacement is perpendicular to the direction of wave propagation. > Vibrates at 90 degrees to the direction the wave is moving > Particles of the medium vibrate up and down perpendicular to the direction of the wave. Transverse Wave Direction of waveThe hand moves the string up and then returns, producing a transverse wave. (a) Transverse wave on a string »/ ‘Motion of the wave wom As the wave passes, each panicle of the string moves up and then down, transversely to the motion ofthe wave itself dransverse Wave qrevatcage-l | ypes of Mechianicalwaves b. Longitudinal wave > A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction that the wave moves. > a wave that causes particles to vibrate in the same direction as (parallel to) the direction that the wave is traveling. > particles of the medium vibrate back and forth parallel to the direction of the wave. Longitudinal Wave Particle movement => tn — Direction of wave Compression rarefraction (ooo andoodancoddThe piston moves to the right, compressing the gas or liquid, and then returns, producing a longitudinal wave. (©) Longitudinal wave in a fluid Particles of the fluid _ Aste wave pases, ach <> v voe article of the fluid moves a Corea — forward and then back, parallel to the motion of the wave itself c. Surface waves » Asurface wave is a wave in which particles of the medium undergo a circular motion. » Surface waves are neither longitudinal nor transverse. > Ina surface wave, itis only the particles at the surface of the medium that undergo the circular motion,The board moves to the right and then returns, producing a combination of longitudinal and transverse waves. (0) Waves on the surface of aliquid Surface particles ofthe liquid AAs the wave passes. each particle ofthe liquid surface » A periodic wave is a series of regular timed disturbances in a medium. > series of evenly timed disturbances in a medium. » For a periodic wave, the shape of the string at any instant is a repeating pattern. » A periodic wave generally follows a sine wave pattern, ON eS Wavelength * Pulse S/\_ Periodic Wave APSAMotion of the wave Amplitude A Trough Amplitude A The SHM of the spring and mass generates a sinusoidal wave in the string. Each particle in the string exhibits the same harmonic motion as the spring and mass; the amplitude of the wave is the amplitude of this motion. > Wavelength of the wave, denoted by 2. > The wave patter travels with constant speed and advances a distance of one wavelength }. in a time interval of one period 7. > So the wave speed is given by v= A/T because f = 1/T, v=Af (periodic wave) The speed of propagation equals the product of wavelength and frequency. The frequency is a property of the entire periodic wave because all points on the string oscillate with the same frequency £Question Sound waves are longitudinal waves in air. The speed of sound depends on temperature; at 20°C it is 344 m/s (1130 ft/s). What is the wavelength of a sound wave in air at 20°C if the frequency is 262 Hz (the approximate frequency of middle C on a piano)? IDENTIFY and SET UP: This problem involves Eq. (15.1), v = Af, which relates wave speed v, wavelength A, and frequency f for a periodic wave. The target variable is the wavelength A. We are given v = 344 m/s and f = 262 Hz = 262 st EXECUTE: We solve Eq. (15.1) for A:Using an oscillating piston to make a sinusoidal longitudinal wave in a fluid. creates a compression (a zone of high density); faction (a zone of low density), Forward motion of the plus backward motion creates ar Compression _Rarefaction Plunger oscillating in SHM at eee Wavelength A is the distance between corresponding points on successive cycles. Wave Function for a Sinusoidal Wave ——y > Waves on a string are transverse; during wave motion a particle with equilibrium position x is displaced some distance y in the direction perpendicular to the x-axis. => > The value of y depends on which particle we are talking about (that is, y depends on x) and also on the time ¢ when we look at it. Thus y is a function of both x and t; EE Fy-y Qo y » > We call y(x, /) the wave function that describes the wave. ke | Wavelength A k= > If we know this function for a particular wave motion, we can use it to find the displacement (from equilibrium) of any particle at any time.» Let’s see how to determine the form of the wave function for a sinusoidal wave. » Suppose a sinusoidal wave travels from left to right (the direction of increasing x) along the string, Oscillator Thre points onthe string, generating wave one halfwavelength apart > Every particle of the string oscillates with simple 1 harmonic motion with the same amplitude and Le frequency. ce > The particle at point B in Fig is at its maximum positive y) Pima | Poin tid value of y att=0 ! > Suppose that the displacement of a particle at the left end of the string (x = 0), where the wave originates is viven. by y(x = 0,1) = Acoswt = Acos2zft The wave disturbance travels from x = 0 to some point x to the right of the origin in an amount of time given by x/v, where v is the wave speed. So the motion of point x at time # is the same as the motion of point x = 0 at the earlier time ¢ — x/v. Hence we can find the displacement of point x at time f by simply replacing ¢ in Eq. by (1 — x/v). When we do that, we find the following expression for the wave function: v(x.1) = Aeos|o(+ - *)| Because cos(—@) = cos, we can rewrite the wave function as x , (sinusoidal wave ya 1) = Acos| o(# - ‘)| = Acos|2ar( * = ‘)] moving in +x-direction)Recause cos(—@) = cos@, we can rewrite the wave function as = 1)] = scm[2er(5=2)] seve +hx-direction) en We can rewnite the wave function given by above Eq. in several different but useful forms. ¥( valor xt (sinusoidal wave moving E05 Acoo[an($ =| in +x-direction) It’s convenient to define a quantity k, called the wave number: 2 k= =" (wave number) Substituting A = 277/k and f = w/27 into the wavelength-frequency relation- ship v = Af gives w= vk (periodic wave) We can then rewrite Eq. as (sinusoidal wave moving U5 1) 7A cos( Kx last) ereersirection)Chal Koni Khaxy hai Restoring force returning the system to equilibrium Vo Inertia resisting the return to equilibrium Above Equation tells that the wave speed should increase when the tension F increases but decrease when the mass per unit length increases. > When two or more waves travelling in a medium overlap each other, the net displacement of the medium is equal to algebraic sum of individual displacement of all waves. > FY: Yo Yas Yq are individual displacement of waves then for n waves the resultant displacement will be yeytytyst.. + Yn yO) = yi) + yo 0)[Standingwaves > Standing waves (also known as stationary waves) are set up as a result of the superposition of two waves with the same amplitude and frequency, travelling at the same speed, but in opposite directions. > Forestring-clamped at both end. at certain frequencies the-interference between the forward wave and the reflected wave produces a standing wave pattern A = antinodes: points at whic (@) Swing is one-half wavelength long. ) String is one wavelength long. (© Swing is one and a half wavelengths long. — a @ String is two wavele hs long> Let’s now consider a string of a definite length L, rigidly held at both ends. > let’s first examine what happens when we set up a sinusoidal wave on such a string. The standing wave that results must have a node at both ends of the string. > fa string with length L is fixed at both ends, a standing wave can exist only if its wavelength satisfies Eq L ny (nm = 1,2,3,...) (string fixed at both ends) Solving this equation for A and labeling the possible values of A as Ay, we find An 1,2,3,...) (string fixed at both ends) fue ng = Mf (n = 1,2,3, (string fixed at both ends) > These frequencies are called harmonics, and the scrics is called a harmonic series. } _fris the second harmonic or the first overtone, f; is the third harmonic or the second overtone, and so on. The first four normal modes of a string fixed at both ends (2) 1» = 1: fundamental frequency, f, N (b) n N A N A N —— | ke 24 =, ———_-»| : second harmonic, fy (first overtone)(©) n= 3: third harmonic, f; (second overtone) N A N A N A N (d) n = 4: fourth harmonic, f, (third overtone) N A N AN AN AN the fundamental frequency of a vibrating string is f; = 0/2L. The speed of v waves on the string is determined by Eq. v= VF/p. ve (string fixed at both ends) This is also the fundamental frequency of the sound wave created in the surrounding air by the vibrating string.PROBLEM In an attempt to get your name in Guinness World Records, you build a bass viol with strings of length 5.00 m between fixed points. One string, with linear mass density 40.0 g/m, is tuned to a 20.0-Hz fundamental frequency (the lowest frequency that the human ear can hear). Calculate (a) the tension of this string, (b) the frequency and wavelength on the string of the second harmonic, and (c) the frequency and wavelength on the string of the second overtone. ENECUTE: (a) We solve Eq. (15.35) for F: F = 4yL"f,7= 4(40.0 x 10~* kg/m)(5.00 m)?(20.0 s“')? = 1600 N = 360 Ib (b) From Egs. (15.33) and (15.31), the frequency and wave- length of the second harmonic (n = 2) are f2 = 2fi = 2(20.0 Hz) = 40.0 Hz 00 m) = 5.00m (©) The second overtone is the “second tone over” (above) the fundamental—that is, n = 3. Its frequency and wavelength are fs= fi 2p 2(5.00m) aa = = = 333m 3(20.0 Hz) = 60.0 HzSolve following problems from Chapter 14. Exercise questions Q14.1 Q.14.10 Q14.16 14.53 Solve following problems from Chapter 15. 1) Exercise questions Qis.1 Q15.3 Q15.4 Q1s5.15