Physics in Faculty of

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Leon Tyler
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2 Why we study Physics in Faculty of Engineering?

3 Dimensional analysis Scalars and vector analysis Rotational of a rigid body about a fixed axis Rotational kinematics

4 1. Dimensional analysis The ward dimension (Dim.) denotes the physical nature of a quantity Basic physical quantity Symbol Unit Dimension Length l m, cm, mm L Mass m Kg, g, mg M Time t S, h, day T Derived Physical quantity Law Dimension Velocity (v) Acceleration (a) Force (F) Momentum (P) Work (W) Power (P) Pressure Density ( ) V = distance time a = velocity time F = mass * acceleration = m.a P = mass*velocity W = force*distance P = Work time P = Force Area = Mass Volume L T LT 1 T = L T 1 = L T M L T = M 1 L 1 T M L T 1 = M 1 L 1 T 1 M L T * L = M L T M L T T M L T L = M L T 3 = M 1 L 1 T M L 3 = M1 L 3

5 1. Dimensional analysis Dimensional Analysis: It is a technique used to check the validity of an equation or to assist in deriving an equation. Note: Validation of the mathematical physical equations requires that: Dimensions of the left hand side [L.H.S] = Dimension of the right hand side [R.H.S] Example: 1- show that the expression V = V 0 + at is dimensionally correct. Solution: L.H.S R.H.S [V] = L T 1 [V 0 ] = L T 1 & [a.t] = L T T = L T 1 Dim. of L.H.S = Dim. of R.H.S Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

6 1. Dimensional analysis Example: Using the dimensional analysis, check the validity of the equation x = 1 a t where x is displacement, a is acceleration and t is time Solution: L.H.S x = L & R.H.S 1 a t = LT T = LT 0 = L [L. H. S.] = [R. H. S.] Example: 1-3 The expression is dimensionally correct let for a particle moving in a circle of radius r with uniform velocity v and the acceleration a is given by a r n v m Find the formula of a. (Find the values of n and m) Solution: a r n v m a = k r n v m Dim. of L.H.S = Dim. of R.H.S [a] = [k r n v m ] L T = L n (L T 1 ) m L T = L n L m T m L T = L n+m T m from L 1 = n + m & from T - = -m m = 1 = n + n = -1 a = k r n v m a = k r 1 v a = k v r

7 1. Dimensional analysis Example Using the dimensional analysis, test the validity of the equation: T = π periodic time, = 3.14, L is length and g is acceleration due to gravity. L g, where T is Solution: L.H.S T = T, R.H.S π L g = L LT = [L. H. S.] = [R. H. S.] T = T The expression is correct Example: (try to answer) From the expression F = G m 1m r find the dimension of [G]? Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

8 . Scalars and Vector Analysis Vector and scalar quantities Scalar quantity: is completely specified by a single value with an appropriate unit and has no direction. Examples: volume, mass, speed, temp., electrical potential difference and time intervals. Notes Some scalars are always positive, such as mass and speed. Others, such as temperature, can have either positive or negative values. Vector quantity: is completely specified by a number with an appropriate unit plus a direction. Examples: velocity, displacement, weight and electric field distance

. Scalars and Vector Analysis Components of vector and unit vectors

components A x & A y where A = A X + A y A X = A cos A y = A sin

= A x A tan = A y A x A y = A sin A x = A cos sin = cos = tan =

9 . Scalars and Vector Analysis Components of vector and unit vectors - Any vector in x y plane can be represented by its vector components A x & A y where A = A X + A y A X = A cos A y = A sin Components The magnitude of A is A = A x + A y Notes sin = A y A cos = A x A tan = A y A x A y = A sin A x = A cos sin = cos = tan = opposite side hypotenuse side adjacent side hypotenuse side opposite side adjacent side

. Scalars and Vector Analysis unit vectors - unit vector is dimensionless vector - its magnitude is exactly 1 (1 unit in length) - used to specify a given

+ B X ) i + (A y + B y ) j R= R x i + R y j R x = A X + B X R y = A y + B y R = R x + R y In three dimensions: x, y, z A = A X i + A y j + A Z k & B = B X i + B

10 . Scalars and Vector Analysis unit vectors - unit vector is dimensionless vector - its magnitude is exactly 1 (1 unit in length) - used to specify a given direction A = A X i + A y j Sum of two vectors A = A X i + A y j & B = B X i + B y j The result is vector R R = A + B = (A X i + A y j) + (B X i + B y j) = (A X + B X ) i + (A y + B y ) j R= R x i + R y j R x = A X + B X R y = A y + B y R = R x + R y In three dimensions: x, y, z A = A X i + A y j + A Z k & B = B X i + B y j + B Z K R = A + B = (A X + B X ) i + (A y + B y ) j + (A Z + B Z ) K i,j and k represent unit vectors pointing in the positive x, y, and z directions, respectively.

11 . Scalars and Vector Analysis Example:.3 Find the sum of the two vectors A and B lying in the xy plane and given by Solution: A = i + j & B = i -4J A = A X i + A y j & B = B X i + B y j A = i + j & B = i - 4 J A X =, A y = B X =, B y = -4 R = A + B = (A X + B X ) i + (A y + B y ) j = ( +) i + (-4)j R = 4 i j R x = 4, R y = - R = R x + R y tan = R y R x = 4 + ( ) = = 0 = 4.47 = tan-1 4 = 3330 Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

12 . Scalars and Vector Analysis Example: -4 Find the resultant displacement for Solution: d 1 = i + 3j - k, d = i - j - 3k & d 3 = - i + j cm R = d 1 + d + d 3 = (1+ -1) i + (3-1+1) j + (-1-3+0) k = i + 3 j - 4 k cm R = R x i + R y j + R z k R = R x + R y + R z = ( 4) = = 9 = 5.39 cm Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

13 . Scalars and Vector Analysis Example: -5 A hiker begins a trip by first 5 km south-east, in nd day 40 Km in a direction 60 0 north of east. (a) determine the components of the displacement for 1 st and nd days. solution:- A X = A cos = A cos (-45) = 5 Km *0.707 = 17.7 km A y = A sin = A sin (-45) = - 5 Km *0.707 = Km B X = B cos 60 = 40 *(0.5) = 0 Km B y = B sin 60 = 40 *(0.866) = 34.6 Km (b) determine the rectangular components of the total displacement of the trip. the resultant displacement R = A + B R x = A x + B x = = 37.7 Km R y = A y + B y = = 16.9 Km in unit vector from R is R = (37.7 i j) Km Exercise:- From the previous example, find magnitude and direction of R R = R x + R y = (37. 7) +(16. 9) = Km 16.9 = tan = 4.10 (north of east)

14 Rotation of a rigid body about fixed axis Angular velocity and angular acceleration Rotational kinematics Relationships between angular & linear quantities Rotational kinetic energy

Rotation of a rigid body about fixed axis 10-1: Angular velocity and angular acceleration:- Suppose a rigid body rotates about a fixed axis through O.

15 Rotation of a rigid body about fixed axis 10-1: Angular velocity and angular acceleration:- Suppose a rigid body rotates about a fixed axis through O. the position of the particle is described by polar coordinates (r, ) Notes: angular position (rad)= S r * 1 rad is the angle which has arc length equal to the radius * 1 rad = 360 = 180 = = * (rad) = 180 (deg.) * (rad) = * In linear motion [1 revolution = 360 (deg)] In rotational motion Ex. = 60 deg. = 45 deg. [1 revolution = (rad)] (rad) = 180 *60 = 3 (rad) (rad) = 180 *45 = 4 (rad) Arc length radius S = circumference = r = S/r = r/r = (rad) 1 rad = / y r s 180 x (rad) (deg) angle displacement = - 1 (rad) The angular velocity = 1 t t 1 = Δ Δt = d dt (rad S 1 ) The angular acceleration = 1 t t = Δ 1 Δt = d dt (rad S )

10- Rotational kinematics:- distance Initial velocity Final velocity acceleration Rotational motion 0 Linear motion X V 0 V a First equation Rotational motion = 0 + t Linear motion V = v 0 +a t

16 10- Rotational kinematics:- distance Initial velocity Final velocity acceleration Rotational motion 0 Linear motion X V 0 V a First equation Rotational motion = 0 + t Linear motion V = v 0 +a t Second equation Third equation = t + 1 t = - 0 = 0 t + 1 t = 0 + ( - 0 ) X = x 0 + v 0 t+ 1 a t X = X - x 0 = v 0 t+ 1 a t V = v 0 + a(x-x 0 ) Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

17 Example: 10-1 A wheel rotates with a constant angular acceleration of 3.5 rad/s. If the angular Or velocity of the wheel is rad/s at t o = 0, (a) find through S = t + 1 t = - 0 = 0 t + 1 t = ( rad s ) (S) + 1 = 11 rad = 11 = 1.75 rev. = 11 * 180 = 6300 (b) Find at t = S No. of rev. = 0 = 0 + t = rad S -1 + (3.5 rad S ) (S) = 9.0 rad/s Exercise: 1 (3.5 rad S ) (S) = 11 rad = = 1.75 rev. From the previous example find the angle between t=s and t = 3S = t + 1 t At t = 3S - 0 = 0 t + 1 t = *3 + ½(3.5)(3) = 1.75 rad 3 - = = rev. (rad) 0 rev

10-3 Relationships between angular and linear quantities:- The

The tangential acceleration a t = dv dt = r d dt Tangential acceleration

centripetal acceleration a r = r, = r a r = r r The total linear

18 10-3 Relationships between angular and linear quantities:- The tangential velocity = ds dt = d dt = S r dθ r = r dt Tangential velocity The tangential acceleration a t = dv dt = r d dt Tangential acceleration s = r = r (1) Angular velocity a t = r () Angular acceleration The centripetal acceleration a r = r, = r a r = r r The total linear acceleration a r = r = r (3) a = a t + a r a = a t + a r = r + r 4 = r ( + 4 ) a = r + 4 (4)

19 Relationships between angular and linear quantities:- Example: 10. A turntable of record player rotates at 33 rev./min. and take 0 S to come to rest. (a) what is the angular acceleration? solution: 1 rev. = (rad) 0 (initial angular velocity) = (33 rev. min = 0 + t, 0-0 = t = 0 t = 0, t = 0 S = ) ( rad rev. min. ) (1 60 = rad/s ) = 3.46 rad/s (b) how many rotation does it make before coming to rest? = - 0 = 0 t + 1 t = 3.46*0 + ½ *(-0.173)(0) = 34.6 rad the No. of rev. = (rad) = 34.6 π = 5.5 rev. (c) if the radius is 14 cm, what are the magnitudes of a t and a r at t =0? a t = r, a r = r a t = r = (14 cm)( rad S )= -.4 cm/s a r = r = (14 cm)(3.46 rad S ) = 168 cm/s Exercise 0 = 3.46 rad/s v = r = 14*3.46 = 48.4 cm/s rev. (rad)

10-4 Rotational kinetic energy:- For a rigid body rotates about z axis with, the mass of i th particle is m i and its speed is v i, the kinetic energy of this particle is K i = ½ m i v i The total

20 10-4 Rotational kinetic energy:- For a rigid body rotates about z axis with, the mass of i th particle is m i and its speed is v i, the kinetic energy of this particle is K i = ½ m i v i The total kinetic energy of the body is the sum of the kinetic energies of the individual particles K = K i = ½ m i i, i = r i = ½ m i r i = ½ ( m i r i ) But I = m i r i (the moment of inertia) The kinetic energy of rotating rigid body K is K = ½ I J Kg/m rad/s Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

Rotational kinetic energy:- Exercise: 10.3 The oxygen molecule Let the diatomic molecule oxygen O rotates in xy plane about z axis (a) If m for an oxygen atom is.

21 Rotational kinetic energy:- Exercise: 10.3 The oxygen molecule Let the diatomic molecule oxygen O rotates in xy plane about z axis (a) If m for an oxygen atom is.66*10-6 Kg, find the moment of inertia I about z axis. O d y d = 1.1*10-10 m d O x I = m i r i I = m (d/) + m (d/) = md I= 1.95*10-46 Kg.m = (.66x10 6 ) (1.1x10-10 ) (a) If the molecule about z axis is.0*10 1 rad/s, find the rotational kinetic energy K = ½ I = ½ (1.95*10-46 )(*10 1 rad/s) = 3.89*10 - j Note: The translational kinetic energy is about 6.*10-1 j K rot. k trans. by one order Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

Rotational kinetic energy:- Exercise: 10.4 : Four rotating particles Four point masses as in fig. (a) If the rotation about y axis with, find the I y and k about y axis.

22 Rotational kinetic energy:- Exercise: 10.4 : Four rotating particles Four point masses as in fig. (a) If the rotation about y axis with, find the I y and k about y axis. Note: the masses m lie on y axis has no I y where r i = 0 I y = m i r i = Ma + Ma = Ma k = ½ I y = ½ (Ma ) k= Ma (the mass m is not included) (b) If the system rotate in xy plane about z axis, find the I z and k about z axis. I z = m i r i = Ma + Ma + m b + m b = Ma + mb k = ½ I z = ½ (Ma + mb ) k = (Ma + mb ) (all masses included) Notes: Not all masses K y K z I y I z All masses included less work is required to rotate about y axis than that required to rotate about z axis

Moments of inertia for objects of different shapes: For a uniform

plane of the hoop through its center is MR.

perpendicular to the rod passing through its center of mass is

23 Moments of inertia for objects of different shapes: For a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop through its center is MR. For a uniform rigid rod of length L and mass M about an axis perpendicular to the rod passing through its center of mass is (1/1) ML. For a uniform solid cylinder has a radius R, mass M, and length L about its axis is (1/)MR.

Summary of important laws angle displacement = - 1 (rad) angular velocity = 1 t t 1 = Δ Δt = d dt (rad S 1 ) angular acceleration = 1 t t 1 = Δ Δt = d dt (rad S ) The tangential velocity = r The

24 Summary of important laws angle displacement = - 1 (rad) angular velocity = 1 t t 1 = Δ Δt = d dt (rad S 1 ) angular acceleration = 1 t t 1 = Δ Δt = d dt (rad S ) The tangential velocity = r The tangential acceleration r v Rotational motion = 0 + t = t + 1 t = 0 t + 1 t Linear motion V = v 0 +a t X = x 0 + v a t X = X - x 0 = v a t a t = r r a t = 0 + ( - 0 ) V = v 0 + (x-x 0 ) The centripetal acceleration The moment of inertia The moment of inertia a r = r = r a r r I = m i r i K = ½ I The total linear acceleration a = r + 4 a r + 4 (rad) rev. 0 rev (rad) (deg)

25 Static equilibrium and elasticity: Elastic properties of solids 1- Young s Modulus - Shear Modulus 3- Bulk Modulus

1-4 Elastic properties of solids Elastic modulus = Stress Strain Force per unit area causing deformation A measure of degree of deformation 1- Young s modulus: it measures the resistance of solid to

26 1-4 Elastic properties of solids Elastic modulus = Stress Strain Force per unit area causing deformation A measure of degree of deformation 1- Young s modulus: it measures the resistance of solid to change its length. - Shear modulus: it measures the resistance of the sliding of the planes of solid on each other. 3- Bulk modulus: it measures the resistance of solids or liquids to change their volumes. 1- Young modulus: (Elasticity of length) Stress-strain curve y = tensile stress = F/A (N/m ) tensile strain L/L 0 1-Firstly, the relationship is a straight line (i.e. Hook s law)and the object will return to its original shape after the stress is removed. - Beyond the elastic limit, the relationship will be not a straight line and the object is permanently distorted and will not return to its original shape after the stress is removed. 3- As the stress is increased even further, the material will break.

- Shear modulus: (Elasticity of shape) x x S = shear stress = F/A shear strain x/h (N/m ) Note: The force is parallel and there is

27 1-4 Elastic properties of solids Elastic limit: The maximum stress that can be applied to an object before it becomes permanently deformed. - Shear modulus: (Elasticity of shape) x x S = shear stress = F/A shear strain x/h (N/m ) Note: The force is parallel and there is no change in volume h A F 3- Bulk modulus: (volume of length) B = - Volume stress Volume strain = - F/A = - P V/V V/V (N/m ) Note: Negative sign mean that stress 1/v (if stress increases then v decreases) Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

28 1-4 Elastic properties of solids Ex. 1.6 A load of 10 kg is supported by a wire of length m and cross-sectional area 0.1 cm. The wire is stretched by 0. cm. Find the tensile stress, tensile strain and Young s modulus for the wire from this information. (g = 9.8 m/s ) Solution: a- tensile stress = F/A = mg/a= = 1*108 N/m L = m b- tensile strain= L/L 0 = = 0.11*10 - c- y = tensile stress tensile strain = = *1010 N/m m.g L= 0. cm m = 10 kg Ex.1.7 A solid lead sphere of volume 0.5 m 3 is lowered to a depth in the ocean where the water pressure is equal to 10 7 N/m. The bulk modulus of lead is equal to N/m. What is the change in volume of the sphere? Solution: B = - v = - V P B Volume stress Volume strain = - F/A V/V =- P V/V = = -1.3 * 10-3 m 3 Note: The negative sign indicates a decrease in volume.

29 The Heat 1- Heat capacity - Specific heat 3- Heat transfer

30 (0-): Heat capacity specific heat Heat capacity: C it is the amount of heat energy needed to raise the temp. of the substance by one degree (1c 0 ) Q = C T C = Q T (J/ 0 C) Specific heat: c it is the amount of heat energy needed to raise the temp. of 1kg of the substance by one degree (1c 0 ) c = C m C = c.m S.H Q = m.c. T j kg Jkg -1 c 0-1 c 0 Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

(0-): Heat capacity & specific heat Measuring specific heat Hot substance cold water Mixture (substance+water) After mixing Mass m x m w Specific heat c x c w Final temp.

31 (0-): Heat capacity & specific heat Measuring specific heat Hot substance cold water Mixture (substance+water) After mixing Mass m x m w Specific heat c x c w Final temp. = T f Initial temp T x T w Heat loss by substance = Heat gain by water m x c x T = m w c w T m x c x (T x T f ) = m W c W (T f T W ) C X = m W cw (T f TW) m x (Tx T f ) Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

32 (0-): Heat capacity & specific heat Ex. 0. m x = 0.05 kg at T x = 00 c 0, m w = 0.4 kg at T w = 0 c 0, T f =.4 c 0. Find c x? Solution: Q loss = Q gain m x c x (T x T f ) = m W c W (T f T W ) (0.05) c x (00-.4) = (0.4) (4186 j/kg c 0 ) (.4-0) c x = 453 j/kg c 0 Exercise 1 : what is the total heat transferred to the water? Solution: Q = mc T = 0.4 * 4186 * (.4 0) Q = 400 j Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

33 (0-): Heat capacity & specific heat Ex. 0.3 A bullet of Ag with mass m = g, velocity v = 00 m/s. what is the temp. change of the bullet? Solution: K.E of the bullet = ½ mv = ½ ( *10-3 ) (00) = 40 j All K.E is transferred into heat K.E = Q = m c T T = Q mc = 40 ( 10 3 ) (34 j kg 1 c 0 1 ) = 85.5 c0 Exercise : if the bullet is lead with the same mass and velocity. Find T? Solution T = 156 C 0 Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

0-7: Heat transfer 1- Heat conduction: it takes place in solid -

radiation: it is the emission of electromagnetic waves (EMW) from

34 0-7: Heat transfer 1- Heat conduction: it takes place in solid - Heat convection: it takes place in fluids (liquids & gases) 3- Heat radiation: it is the emission of electromagnetic waves (EMW) from the hot body Heat conduction: the rate of heat flow H = Q T H A heat time (j/s watt) H T H A T x H 1 x H = - k A T x H Note: T T 1 T increase Negative sign means that the direction of heat flow H opposite to the direction of temp. gradient dt dx

0-7: Heat transfer H = k A (T T 1 ) L Temp. diff.

and thermal conductivities K 1, K, K 3,.

35 0-7: Heat transfer H = k A (T T 1 ) L Temp. diff. (c 0 ) length (m ) Thermal conductivity (w/m o c ) for compound slab of several materials of thicknesses L 1, L, L 3,. and thermal conductivities K 1, K, K 3,.. Cross-section area (m ) L - Heat convection H = A(T T 1 ) i L i K i It is the transfer of heat in fluids by the motion of the fluid itself. The figure shows that, the liquid near the heat source is heated and expands slightly, becoming lighter than the overlying cooler fluid. It then rises and it is replaced by cooler, heavier fluid. When the warmer fluid arrives at the cooler region of the container, it cools, and begins to sink again, and so on. When, the container completely becomes warm, the convection process may be stopped. H = q A T Temp. diff. convective heat transfer constant

36 0-7: Heat transfer 3- Heat radiation All objects radiate energy continuously in the form of electromagnetic waves The rate at which an object radiates energy is proportional to the fourth power of its absolute temperature. This is known as Stefan s law Stefan s law H = e σ A T 4 Is called emissivity Its value between 0-1 It is Stefan s constant σ = 5.76 x 10-8 W m -- K -4 Important Facts The value of emissivity depends on the surface of the object A shiny surface has a small value of emissivity, while a black one has e near to 1 A good emitter must also be a good absorber A good reflector is poor emitter & has small value of emissivity. A perfect absorber (emitter) has emissivity =1 Any object that absorbs all incident radiation appears black

37 0-7: Heat transfer Example: 0.9 Find T ; the temp. at the interface and H Solution: H 1 H H

38 Sound waves 17-1 velocity of sound wave 17- Harmonic sound waves 17-3 Energy & intensity of harmonic sound waves

39 17-1: Velocity of sound waves -Sound wave is longitudinal waves travel through a compressible medium. - As the waves travel through a medium, the elements of the medium vibrate longitudinally to produce changes in density and pressure. compression - The speed of sound waves in a medium depends on.. a- compressibility of medium b- density of medium When a piston is pushed to right, the gas compressed and the compressed region continuous to move to right with speed pulses v rarefaction Compressed regions The speed of sound wave in gases The speed of sound wave in solids Bulk modulus B v = Equilibrium density Young s modulus ρ Y v = Equilibrium density ρ The speed of mechanical waves in general form is v = elastic property internal property Note: v air v liq. v solids

17-1: Velocity of sound waves Example 17.1 (velocity of sound in solid) If a solid bar is struck at one end with a hammer, a longitudinal pulse will propagate down the bar with a speed v.

40 17-1: Velocity of sound waves Example 17.1 (velocity of sound in solid) If a solid bar is struck at one end with a hammer, a longitudinal pulse will propagate down the bar with a speed v. Find the speed of sound in an aluminum bar if Young s modulus (Y = N/m and density = kg/m 3 ). Solution: v = Y = ρ = 509 m/s Al bar Example 17.1 (velocity of sound in liquid) Find the speed of sound in water, which has a bulk modulus of about B = N/m and a density of about = kg/m 3. Solution: v = B = ρ = 1449 m/s Note: v air = 343 m/s, v liq. = 1493 m/s & v solids = 5130 m/s

- The position of a small element s(x,t) S(x,t) = S m cos (k x - t) S m : maximum displacement K : wave number (k = π

41 17-: Harmonic sound waves - When a piston oscillate sinusoidaly compression & rarefaction are continuously set up, any small element of medium moves with Simple harmonic motion parallel to the direction of the wave. - The position of a small element s(x,t) S(x,t) = S m cos (k x - t) S m : maximum displacement K : wave number (k = π ) : angular frequency t : time - The variation in pressure P is also harmonic P = P m sin (kx - t) Maximum change in pressure P m = v S m

42 17-3: Energy and intensity of harmonic sound waves Notes: Power of sound wave (watt) = ½ A v ( S m ) Intensity of sound (w/m ) = I = power area = ½ v ( S m) Threshold of hearing I = 10-1 w/m Threshold of pain I = 1 w/m Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

43 Superpostion & standing waves 18- Standing wave 18-5 Standing wave in air column

44 antinode 18-: Standing waves * Standing wave occurs due to the superposition of two waves have the same frequency traveling in opposite directions in medium. * consider two sinusoidal waves traveling in opposite direction in the same medium * The wave traveling to the right (+ x direction) is y 1 = A 0 sin (kx - t) * The wave traveling to the left (- x direction) is (A 0 is the amplitude) x= n 4 y 1 = A 0 sin (kx + t) * Adding the two waves gives the resultant wave y = y 1 + y = A 0 sin (kx - t) +A 0 sin (kx - t) x= node 3 4 y = (A 0 sin kx) cos t (Standing wave function) Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

45 18-: Standing waves Amplitude of standing wave Maximum A 0 Sin kx = max. if Sin kx =1 ( A 0 Sin kx) Minimum A 0 Sin kx = min. If Sin kx = x= n 4 Kx = π, 3π But k = π π x = π, 5π x = 4 Position of antinodes Kx = π, π, 3 π But k = π π x = π x = Position of nodes A 0 x= / 3 4 A A A A N N N N N x = 4, 3 4, 5 4, n 4 x =,, 3, n A A A A / Notes Node: The position at which the sound waves are not heard Antinode: The position at which the sound waves are hear Wavelength of standing wave: The double of distance between two successive nodes or antinodes.

46 18-: Standing waves Example: 18. Two waves traveling in opposite directions produce a standing wave. The individual wave functions are given by: y 1 = 4 cm si n 3x t and y = (4 cm)sin (3x + t) (a) Find the maximum displacement of the motion at x =.3 cm. (b) Find the positions of eh node and antinodes. Solution: ( x and y are in cm). y 1 = A 0 sin (kx ωt) A 0 = 4 cm, k = 3 cm -1, = rad/s y = A 0 sinkx cosωt = 4 sin3x cost = 8sin3x cost The maximum displacement at x =.3 cm y = 8sin3x cost = 8sin 3. 3 = 8sin 6. 9 rad = cm The position of antinodes: x = nλ 4 k = π λ = 3 λ = π 3 x = n 4 π 3 = n π 6 The position of nodes: x = nλ x = n π 3 = n π 3 (n = 1, 3, 5,.. ) cm (n = 1, 3, 5,.. ) (n = 1,, 3,.. ) cm (n = 1,, 3,.. )

47 18-5: Standing waves in air columns Types of air column: a- Open air column (open at each end) b- closed air column (closed at one end) Notes: 1- At open end: There is an anti-node. - At closed end: There is always a node. 1- standing longitudinal wave in pipe open at two ends first harmonic: second harmonic: Third harmonic: L = 1 / 1 = L L = / = L L = 6 3 /4 3 = /3 L v = f 1 1 f 1 = v λ 1 = v L v v = f f 1 = = v λ L v v = f 3 3 f 3 = = 3v λ L f = f 1 f 3 = 3f 1 Since all harmonics are present, we can express the natural frequencies of vibration as: f n = n v L = n f 1 (n = 1,, 3, ) v is the speed of sound in air & n is the order of harmonic

48 18-5: Standing waves in air columns - standing longitudinal wave in pipe closed at one end first harmonic: Third harmonic: Fifth harmonic: L = 1 /4 1 = 4L L = 3 3 /4 3 = 4/3L L = 5/4 4 5 = 5/4 L v = f 1 1 f 1 = v λ 1 = v 4L v = f 3 3 f 3 = v = 3v 3 4L v v = f 5 5 f 5 = = 5v λ 5 4L f 3 = 3f 1 f 5 = 5f 1 In a pipe closed at one end, only odd harmonics are present f n = n v 4L = n f 1 (n = 1, 3, 5, ) v is the speed of sound in air & n is the order of harmonic Assoc. Prof. Mohamed Eltabey - Asst. Prof. Ahmed Saleh

49 18-5: Standing waves in air columns Exercise 18.4 A pipe has a length of 1.3 m. (a) Determine the frequencies of the first three harmonics if the pipe is open at each end. Take v = 344 m/s as the speed of sound in air. (b) What are the three frequencies determined in if the pipe is closed at one end. (c) For the case of the open pipe, how many harmonics are present in the normal human hearing range (0 to 0000 Hz)? Solution: (a) The pipe is open at each end: (b) The pipe is closed at one end: f n = n v L = nf 1 (n = 1,, 3, ) f 1 = v L = 344 = 140 Hz 1. 3 f = f 1 = 140 = 80 Hz f 3 = 3f 1 = = 40 Hz f n = n v 4L = nf 1 (n = 1, 3, 5, ) f 1 = v 4L = 344 = 70 Hz f = 3f 1 = 3 70 = 10 Hz f 3 = 5f 1 = 5 70 = 350 Hz (c) f n = nf = n 140 n = = 14 the no. of harmonics 14, but actually only the first harmonics have sufficient amplitude to be heard

PHYS 1114, Lecture 33, April 10 Contents:

PHYS 1114, Lecture 33, April 10 Contents: 1 This class is o cially cancelled, and has been replaced by the common exam Tuesday, April 11, 5:30 PM. A review and Q&A session is scheduled instead during class