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Lecture 21 Sound →Fluids. Standing Waves I A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions.

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Presentation on theme: "Lecture 21 Sound →Fluids. Standing Waves I A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions."— Presentation transcript:

1 Lecture 21 Sound →Fluids

2 Standing Waves I A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions a and b. Let upward motion correspond to positive velocities. When the string is in position b, the instantaneous velocity of points on the string: a b a) is zero everywhere b) is positive everywhere c) is negative everywhere d) depends on the position along the string

3 Observe two points: Just before b Just after b Both points change direction before and after b, so at b all points must have zero velocity. Standing Waves I A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions a and b. Let upward motion correspond to positive velocities. When the string is in position b, the instantaneous velocity of points on the string: a) is zero everywhere b) is positive everywhere c) is negative everywhere d) depends on the position along the string Every point in in SHM, with the amplitude fixed for each position

4 a b c Standing Waves II A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions a and b. Let upward motion correspond to positive velocities. When the string is in position c, the instantaneous velocity of points on the string: a) is zero everywhere b) is positive everywhere c) is negative everywhere d) depends on the position along the string

5 direction depends on the location When the string is flat, all points are moving through the equilibrium position and are therefore at their maximum velocity. However, the direction depends on the location of the point. Some points are moving upward rapidly, and some points are moving downward rapidly. a b c Standing Waves II A string is clamped at both ends and plucked so it vibrates in a standing mode between two extreme positions a and b. Let upward motion correspond to positive velocities. When the string is in position c, the instantaneous velocity of points on the string: a) is zero everywhere b) is positive everywhere c) is negative everywhere d) depends on the position along the string

6 Two waves with close (but not precisely the same) frequencies will create a time-dependent interference Beats

7 If two sounds are very close in frequency, their sum also has a periodic time dependence: f beat = f 1 - f 2 Beats are an interference pattern in time, rather than in space.

8 In tuning a string, a 262-Hz tuning fork is sounded at the same time as the string is plucked. Beats are heard with a frequency of 6 Hz. What is the frequency emitted by the string?

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10 Pair 1Pair 2 a) pair 1 b) pair 2 c) same for both pairs d) impossible to tell by just looking The traces below show beats that occur when two different pairs of waves interfere. For which case is the difference in frequency of the original waves greater?Beats

11 Pair 1Pair 2 difference in frequency f beat = f 2 – f 1. The beat frequency is the difference in frequency between the two waves: f beat = f 2 – f 1. greater beat frequency greater frequency difference Pair 1 has the greater beat frequency (more oscillations in same time period), so pair 1 has the greater frequency difference. a) pair 1 b) pair 2 c) same for both pairs d) impossible to tell by just looking The traces below show beats that occur when two different pairs of waves interfere. For which case is the difference in frequency of the original waves greater?Beats

12 a) depends on the speed of sound in the pipe b) you hear the same frequency c) you hear a higher frequency d) you hear a lower frequency You blow into an open pipe and produce a tone. What happens to the frequency of the tone if you close the end of the pipe and blow into it again? Open and Closed Pipes

13 open pipe of a wave closed pipe of a wavewavelength is larger in the closed pipefrequency will be lower In the open pipe, of a wave “fits” into the pipe, and in the closed pipe, only of a wave fits. Because the wavelength is larger in the closed pipe, the frequency will be lower. a) depends on the speed of sound in the pipe b) you hear the same frequency c) you hear a higher frequency d) you hear a lower frequency You blow into an open pipe and produce a tone. What happens to the frequency of the tone if you close the end of the pipe and blow into it again? Open and Closed Pipes Follow-up: What would you have to do to the pipe to increase the frequency?

14 When Mary talks, she creates an intensity level of 60 dB at your location. Alice talks with the same volume, also giving 60 dB at your location. If both Mary and Alice talk simultaneously from the same spot, what would be the new intensity level that you hear? a) more than 70 dB b) 70 dB c) 66 dB d) 63 dB e) 61 dB Decibel Level

15 When Mary talks, she creates an intensity level of 60 dB at your location. Alice talks with the same volume, also giving 60 dB at your location. If both Mary and Alice talk simultaneously from the same spot, what would be the new intensity level that you hear? With two voices adding up, the intensity increases by a factor of 2, meaning that the intensity level is higher by an amount equal to →  = 10 log(2) = 3 dB. The new intensity level is  = 63 dB. Decibel Level a) more than 70 dB b) 70 dB c) 66 dB d) 63 dB e) 61 dB

16 a) about 10 b) about 12 c) about 60 d) about 108 Workplace Noise A factory floor operates 120 machines of approximately equal loudness. A plant safety inspection shows that the sound intensity level of 101 dB is too high, and must be lowered to 91 dB. How many of the machines, at least, would need to be turned off to bring the sound level into compliance?

17 Fluids

18 Pressure The same force applied over a smaller area results in greater pressure – think of poking a balloon with your finger and then with a needle. Pressure is not the same as force! Pressure is force per unit area Pressure is a useful concept for discussing fluids, because fluids distribute their force over an area

19 Atmospheric Pressure Atmospheric pressure is due to the weight of the atmosphere above us. = 1 pascal (Pa) Pascals pounds per square inch bars Various units to describe pressure:

20 Atmospheric Pressure Atmospheric pressure is due to the weight of the atmosphere above us. Hemi-spheres: ~3 inches radius, ~30 in 2 area ~450 lbs! How much is 1 atm ? mass of quarter ~ 0.0057 kgarea of quarter ~ 3x10 -4 m 2 Pressure from weight of one quarter : 180 N/m 2 To get 101kPa, one must be buried under a stack ~560 quarters, or 14 rolls, deep! 4 in 2 area -> ~60 lbs! Put a 1 atm block on your hand?

21 Density, height, and vertical force How does tension change in a vertical (massive) rope? How does normal force change in stack of blocks? In a fluid, how does force change with vertical height?

22 Density The density of a material is its mass per unit volume:

23 Pressure and Depth Pressure increases with depth in a fluid due to the increasing mass of the fluid above it.

24 Pressure and depth Pressure in a fluid includes pressure on the fluid surface (usually atmospheric pressure)

25 Pressure depends only on depth and external pressure (and not on shape of fluid column)

26 Equilibrium only when pressure is the same Unequal pressure will cause liquid flow: must have same pressure at A and B Oil is less dense, so a taller column of oil is needed to counter a shorter column of water

27 The Barometer A barometer compares the pressure due to the atmosphere to the pressure due to a column of fluid, typically mercury. The mercury column has a vacuum above it, so the only pressure is due to the mercury itself. P atm = ρgh Atmospheric pressure in terms of millimeters of mercury: The barometer equilibrates where the pressure due to the column of mercury is equal to the atmospheric pressure.

28 Measuring Pressure The barometer measures atmospheric pressure vs. liquid height of known density (vacuum above the liquid column) The manometer measures a pressure relative to atmospheric pressure Gauge Pressure ΔP= P - P atm = ρ gh

29 a) greater than P A b) equal to P A c) less than P A The Straw You put a straw into a glass of water, place your finger over the top so that no air can get in or out, and then left the straw from the liquid. You find that the straw retains some liquid. How does the air pressure P in the upper part compare to the atmospheric pressure P A ?

30 P A – P –  g H = 0 Consider the forces acting at the bottom of the straw: P A – P –  g H = 0 This point is in equilibrium, so net force is zero. P = P A –  g H pressure Pless outside pressure P A Thus, P = P A –  g H and so we see that the pressure P inside the straw must be less than the outside pressure P A. H a) greater than P A b) equal to P A c) less than P A The Straw You put a straw into a glass of water, place your finger over the top so that no air can get in or out, and then lift the straw from the liquid. You find that the straw retains some liquid. How does the air pressure P in the upper part compare to the atmospheric pressure P A ?

31 Pascal’s principle An external pressure applied to an enclosed fluid is transmitted to every point within the fluid. Hydraulic lift Assume fluid is “incompressible”

32 Pascal’s principle Hydraulic lift Are we getting “something for nothing”? Assume fluid is “incompressible” so Work in = Work out!

33 Buoyancy A fluid exerts a net upward force on any object it surrounds, called the buoyant force. This force is due to the increased pressure at the bottom of the object compared to the top. Consider a cube with sides = L

34 Buoyant Force When a Volume V is Submerged in a Fluid of Density ρ fluid F b = ρ fluid gV Archimedes’ Principle Archimedes’ Principle: An object completely immersed in a fluid experiences an upward buoyant force equal in magnitude to the weight of fluid displaced by the object.

35 Applications of Archimedes’ Principle An object floats when it displaces an amount of fluid equal to its weight. equivalent mass of water wood block equivalent mass of water brass block

36 Applications of Archimedes’ Principle An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water. An object floats when it displaces an amount of fluid equal to its weight.

37 Applications of Archimedes’ Principle The fraction of an object that is submerged when it is floating depends on the densities of the object and of the fluid.

38 Measuring the Density Get the volume from ( T 1 - T 2 ) = V( ρ water g) Get the mass from W = T 1 = mg The King must know: is his crown true gold?

39 On Golden Pond a) rises b) drops c) remains the same d) depends on the size of the gold A boat carrying a large chunk of gold is floating on a lake. The chunk is then thrown overboard and sinks. What happens to the water level in the lake (with respect to the shore)?

40 a) rises b) drops c) remains the same d) depends on the size of the gold weightrequire a lot of displaced water only displaces its volume in water Initially the chunk of gold “floats” by sitting in the boat. The buoyant force is equal to the weight of the gold, and this will require a lot of displaced water to equal the weight of the gold. When thrown overboard, the gold sinks and only displaces its volume in water. This is not so much water—certainly less than before—and so the water level in the lake will drop. On Golden Pond A boat carrying a large chunk of gold is floating on a lake. The chunk is then thrown overboard and sinks. What happens to the water level in the lake (with respect to the shore)?

41 Wood in Water Two beakers are filled to the brim with water. A wooden block is placed in the beaker 2 so it floats. (Some of the water will overflow the beaker and off the scale). Both beakers are then weighed. Which scale reads a larger weight? a b c same for both

42 displaces an amount of water equal to its weight weight of the overflowed water is equal to the weight of the blockbeaker in 2 has the same weight as that in 1 The block in 2 displaces an amount of water equal to its weight, because it is floating. That means that the weight of the overflowed water is equal to the weight of the block, and so the beaker in 2 has the same weight as that in 1. Wood in Water Two beakers are filled to the brim with water. A wooden block is placed in the beaker 2 so it floats. (Some of the water will overflow the beaker and off the scale). Both beakers are then weighed. Which scale reads a larger weight? a b c same for both

43 Wood in Water II A block of wood floats in a container of water as shown on the right. On the Moon, how would the same block of wood float in the container of water? Earth Moon abc

44 weight of water equal to the object’s weight less weight also has less weight A floating object displaces a weight of water equal to the object’s weight. On the Moon, the wooden block has less weight, but the water itself also has less weight. Wood in Water II A block of wood floats in a container of water as shown on the right. On the Moon, how would the same block of wood float in the container of water? Moon abc Earth

45 A wooden block is held at the bottom of a bucket filled with water. The system is then dropped into free fall, at the same time the force pushing the block down is also removed. What will happen to the block? a) the block will float to the surface. b) the block will stay where it is. c) the block will oscillate between the surface and the bottom of the bucket

46 A wooden block of cross-sectional area A, height H, and density ρ 1 floats in a fluid of density ρ f. If the block is displaced downward and then released, it will oscillate with simple harmonic motion. Find the period of its motion. h

47 A wooden block of cross-sectional area A, height H, and density ρ 1 floats in a fluid of density ρ f. If the block is displaced downward and then released, it will oscillate with simple harmonic motion. Find the period of its motion. Vertical force: F y = (hA)g ρ f - (HA)g ρ 1 h at equilibrium: h 0 = H ρ 1 / ρ f Total restoring force: F y = -(Ag ρ f )y h = h 0 - y Analogous to mass on a spring, with κ = Agρ f


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