1 Waves & Sound

2 Objectives FCAT –Periodicity of waves –Movement of particles in transverse vs longitudinal wave

3 Sunshine State Standards SC.H Students know that technological problems often create a demand for new scientific knowledge and that new technologies make it possible for scientists to extend their research in a way that advances science. SC.H Students know that scientific knowledge is used by those who engage in design and technology to solve practical problems, taking human values and limitations into account

4 Chapters 14 & 15 Waves are periodic disturbances that propagate through a medium or space –a medium does not travel with the wave Mechanical waves require a medium Electromagnetic waves do not require a medium

5 Key Formulae Periodic Motion F = -kx PE = ½ kx 2 T = 2  L/g Waves f = 1/T v = f I = P/A  = 10 log I f d = f s v + v d I o v + v s V T in o C = m/s + (0.6m/s o C)(  T)

6 Assignments P&P14 Waves 1-3,6,7,9,10,15-19,31,33,35,40,41,51,52,79 P&P15 Sound 1-3,6-8,12-14,

7 Periodic Motion is repetitive motion Simple harmonic motion is the result of a restoring force on an object being directly proportional to the object’s displacement from equilibrium. That type of force obeys Hooke’s Law: F = -kx F, force, n k, spring constant, n/m X, distance, m Ignore the “–” (means restoring force working against applied force)

8 Sample Problem Page How much force is necessary to stretch a spring 0.25 m when the spring constant is 95 N/m? F = kx F = (95 N/m)(0.25m) F = 24 N

9 Pendulums T, period, sec L, length of pendulum, m g, gravity, 9.8 m/s 2 on Earth What is the period on Earth of a pendulum with a length of 1.0 m? = 2  sr(1.0m / 9.8 m/s 2 ) = 2.0s

10 Measuring Waves, (lambda), wavelength, m v, speed, m/s f, frequency, 1/s, hertz f = 1/T f, frequency, 1/s T, period, s

11 p A sound wave produced by a clock chime is heard 515 m away 1.50 s later. a.What is the speed of sound of the clock’s chime in air? V = d/t = 515 m / 1.50 s = 343 m/s b.The sound wave has a frequency of 436 Hz. What is the period of the wave? t = 1/f = 1/436 Hz = 2.29 x s c.What is the wave’s wavelength?  = V / f = 343 m/2 / 436 Hz = m

12 More on waves: Longitudinal waves – particles of the medium move parallel to the direction of the wave Sound is an example of this type of wave. Sound also requires a medium, categorizing it as a mechanical wave.

13 Transverse Waves Transverse waves – displacement of the particles of the medium are perpendicular to the direction of propagation of the wave.

14 Phase of a wave physics.mtsu.edu

15 Frequency

16 Properties of Waves 1. Rectilinear propagation – advancement of a wave is perpendicular to the wave front 2. Reflection – waves bounce off barriers and rebound in opposite direction –Law of reflection: Incident angle = reflected angle (i = r) 3. Refraction – bending (changing direction) of a wave as it travels from one medium into another

17 More properties… 4. Diffraction – spreading of a wave as it passes beyond the edge of a barrier 5. Interference – result of 2 or more waves passing through the same medium at the same time

Refraction – bending after passing from one substance into another 18 Source: chemicalparadigms.wikispaces.com Source: matter.org.uk

Diffraction – spreading after passing by a barrier 19 geographyfieldwork.com Source: newgeology.us

Interference 20 Source: discoverhover.org Constructive – in phase, increase of amplitude, sounds get louder Destructive – out of phase, cancel one another out

21 Math associated with waves Frequency = 1 / Period (f = 1/T) Period = 1 / Frequency i.e., f = 60cycles per second = 60 hz = 60/s Also:v = f where velocity = frequency x wavelength

22 Sound Waves

23 Math and waves I = P/A Intensity = Power/Area Units: watt / cm 2 Intensity relates to loudness

24 More math…  = 10 log I I o w/cm 2  Intensity level, decibels I, Intensity, w/cm 2 I o, threshhold of hearing, w/cm 2 We all don’t hear the same, so this is a comparative measurement in decibels

25 Intensity Problem Sound energy is radiated uniformly in all directions from a small source at a rate of 1.2 watts. A) What is the intensity of the sound at a point 25 m from the source? Assume no energy is lost in transmission over a spherical area with radius of 25 m. P = 1.2 w r = 25 m

26 Finding the intensity, I I =P = P A4  r 2 =1.2 w 4  (2500 cm) 2 = 1.5 x w/cm 2 Source: hyperphysics.phy-astr.gsu.edu

27 Find intensity level,   = 10 log I I o = 10 log 1.5 x w/cm w/cm 2  =10 log (1.5x10 8 )  = 82 db

28 Intensity Level to Intensity Let’s say that the intensity level of a sound is 25.3 dB. What is the intensity of the sound in w/cm 2 ?  = 10 log I / Io 25.3 dB = 10 log (I/ w/cm 2 ) 2.53 = log I – log log = log I 2.53 – 16 = log I

dB = 10 log (I/ w/cm 2 ) 2.53 = log I – log log = log I  what power do you raise 10 to, to get ? 2.53 – 16 = log I Adding on the left  = log I Raise 10 to the power by this sequence: 2 nd 10 x (-13.5)  3.16 x w/cm 2 = I

30 More to come… Doppler Effect is the apparent change in frequency as a result of relative motion between the source of a sound and a detector. f d = f s v v d v v s f d, frequency heard by detector f s, frequency of source v, velocity of sound in air v d, velocity of detector v s, velocity of source Source: onlinephys.com

31 More on Doppler Stationary detector: Source moving away: v+v s on bottom Source moving toward: v-v s on bottom Stationary source: detector moving toward: v+v d on top detector moving away: v-v d on top

32 Velocity of sound at various temps. * V T in o C = m/s + (0.6m/s o C)(  T) m/s is speed of sound in air at 0 o C 0.6m/s o C rate of change of speed per o C change What is the speed of sound in air at o C? #1  V 10 o C = 331.5m/s +(0.6m/s o C)(10 o C) V = m/s #2  V -5.2 o C = m/s + (0.6m/s o C)(  C) V = = 328 m/s *Notice the 2 constants in red.

33 Doppler Example A stationary civil defense siren has a frequency of 1000 Hz. What frequency will be heard by drivers of cars moving at 15 m/s? A) away from the siren? B) toward the siren? The velocity of sound in air is 344 m/s.

34 Solution: Moving away from the siren f d = f s v - v d = v f d = (1000 Hz) (344 m/s - 15 m/s) 344 m/s f d = 956 Hz Apparent frequency heard by the detector decreases

35 Solution: Moving toward the siren f d = f s v + v d = v f d = (1000 Hz) (344 m/s + 15 m/s) 344 m/s f d = 1044 Hz Apparent frequency increases

36 One More… A police car with a 1000 Hz siren is moving at 15 m/s. What frequency is heard by a stationary listener when the police car is a) receding from the detector? b) approaching the detector?

37 f d = f s v v + v s f d = (1000 Hz) 344 m/s = 958 Hz 344 m/s + 15 m/s P. car moving away f d = f s v v - v s f d = (1000 Hz) 344 m/s = 1046 Hz 344 m/s - 15 m/s P. car coming toward