Sound 14-5 The Doppler Effect © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect Consider the ice-cream truck that cruises through the neighborhood playing that interminable music. Symmetric cross-section Planar cross-section © 2006 By Timothy K. Lund We’ll use a bell for our sound source, and assume it is ringing at a fixed frequency. The sound waves radiate spherically from the bell, as shown in the planar cross-section. Since the ice cream truck is not yet moving, the wave fronts are spherically symmetric.

Sound 14-5 The Doppler Effect We can represent a sound wave as a sine wave - just what you’d see on water if you dropped in a stone. Dobson © 2006 By Timothy K. Lund Whether Dobson is in front of the truck, or behind it, he hears the same frequency. FYI: Sound waves are not, strictly speaking, crests and troughs, like water waves. Sound waves are really alternating continuums of compressed and rarefied air.

Sound 14-5 The Doppler Effect Suppose the ice-cream truck is now moving, but ringing the bell at the same rate as before. © 2006 By Timothy K. Lund Note how the wave fronts bunch up in the front, and separate in the back. The reason this happens is that the truck moves forward a little bit during each successive spherical wave emission.

Sound 14-5 The Doppler Effect If Dobson is in front of the moving truck, he will hear a different frequency than if he is behind it. © 2006 By Timothy K. Lund In fact, as the truck approaches him he will hear a higher than actual frequency. As the truck recedes, he will hear a lower than actual frequency. We call this phenomenon the Doppler effect.

Sound 14-5 The Doppler Effect Now we want to look at TWO successive pulses emitted by a source that is MOVING at speed vs: s = vT Pulse 1 t = 0 Pulse 2 t = T Pulse 1 t = T VS o d = vsT © 2006 By Timothy K. Lund The distance between Pulse 1 at t = 0 and Pulse 1 at t = T is given by s = vT. The distance the Source has traveled in time T is given by d = vsT. This is the distance shown here: The wavelength of the sound heard by the observer Dobson is represented with the symbol o. Note that o = s - d. Thus o = vT - vsT

Sound 14-5 The Doppler Effect FYI: The denominator is less than the numerator so that fo > fs as expected. we can do a little manipulation to get the following result: From o = vT - vsT o = vT - vsT o = (v - vs)T But the frequency perceived by the observer, fo is related to the speed of sound by v = ofo so that o = v fo . But fs = 1/T so that o = (v - vs)T v fo 1 fs = (v - vs) © 2006 By Timothy K. Lund 1 fo (v - vs) v 1 fs = Doppler Effect Source moving TOWARD a stationary observer v v - vs fo = fs 1 1 - fo = fs vs v

Sound 14-5 The Doppler Effect If the source is moving AWAY from the observer, a little thought should convince you that Doppler Effect Source moving AWAY FROM a stationary observer v v + vs fo = fs 1 1 + fo = fs vs v FYI: The denominator is more than the numerator so that fo < fs as expected. © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect FYI: The multiplier is less than 1 so that fo < fs as expected. FYI: The multiplier is greater than 1 so that fo > fs as expected. Sound 14-5 The Doppler Effect If the source is stationary, and the observer is moving toward the source, the effective speed of the emitted wavefronts will be ve = v + vo. Then the frequency heard by the observer is given by ve s v + vo s v + vo v fo = = = fs Doppler Effect Observer moving TOWARD a stationary source v + vo v vo v fo = fs = 1 + fs If the source is stationary, and the observer is moving away from the source, the effective speed of the emitted wavefronts will be ve = v - vo. © 2006 By Timothy K. Lund Then the frequency heard by the observer is given by ve s v - vo s v - vo v fo = = = fs Doppler Effect Observer moving AWAY FROM a stationary source v - vo v fs vo v fo = = 1 - fs

Sound 14-5 The Doppler Effect Of course, there is always the situation where both source and observer are moving. Dare we talk about this case? Well, it turns out that the following formula works for ALL situations: The Doppler Effect v  vo v vs + fo = fs source and observer moving (toward means greater) v v vs + If vo = 0 the formula reduces to © 2006 By Timothy K. Lund fo = fs. v  vo v If vs = 0 the formula reduces to fo = fs.

Sound 14-5 The Doppler Effect FYI: The source is moving TOWARD the observer, so we want a GREATER frequency: USE THE NEGATIVE. Sound 14-5 The Doppler Effect Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is moving along at 25 m/s. (a) What frequency will Dobson hear if he is standing in front of the truck? Assume the speed of sound is 345 m/s. We can use our general formula, with vo = 0: The Doppler Effect v  vo v vs + fo = fs source and observer moving (toward means greater) © 2006 By Timothy K. Lund v v vs + v v - vs fo = fs = fs 345 345 - 25 = ·120 = 129 Hz

Sound 14-5 The Doppler Effect FYI: The source is moving AWAY FROM the observer, so we want a SMALLER frequency: USE THE POSITIVE. Sound 14-5 The Doppler Effect Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is moving along at 25 m/s. (b) What frequency will Dobson hear if he is standing behind the truck? Assume the speed of sound is 345 m/s. We can use our general formula, with vo = 0: v  vo v vs + v v vs + fo = fs = fs v v + vs © 2006 By Timothy K. Lund = fs 345 345 + 25 = ·120 = 112 Hz

Sound 14-5 The Doppler Effect FYI: The source is moving TOWARD the observer, so we want a HIGHER frequency: USE THE POSITIVE. Sound 14-5 The Doppler Effect Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is parked on the side of the road.  What frequency will Dobson hear if he is approaching the truck at 25 m/s on his motorcycle? Assume the speed of sound is 345 m/s. We can use our general formula, with vs = 0: v  vo v v + vo v fo = fs = fs 345 + 25 345 = ·120 © 2006 By Timothy K. Lund = 129 Hz

Sound 14-5 The Doppler Effect FYI: The source and observer are moving TOWARD each other, so we want a HIGHER frequency: USE THE POSITIVE in the top and the NEGATIVE in the bottom. Sound 14-5 The Doppler Effect Suppose the ice cream truck is ringing the bell with a frequency of 120 Hz (about the frequency of a fire alarm) and is traveling at 25 m/s.  What frequency will Dobson hear if he is approaching the truck from the front at 25 m/s on his motorcycle? Assume the speed of sound is 345 m/s. We can use our general formula: v  vo v vs + v + vo v - vs fo = fs = fs 345 + 25 345 - 25 = ·120 © 2006 By Timothy K. Lund = 139 Hz

Sound 14-5 The Doppler Effect MACH SPEEDS If the sound source moves faster than the speed of sound, the source passes up the wave front, producing what is called a sonic boom. © 2006 By Timothy K. Lund vs = 0 vs < v vs = v vs > v Subsonic Mach 1 Supersonic FYI: At subsonic speeds, the wave fronts never cross each other, and thus cannot constructively interfere. FYI: At supersonic speeds, the wave fronts DO cross each other, and thus CAN constructively interfere.

Sound 14-5 The Doppler Effect MACH SPEEDS If atmospheric conditions are just right, evidence of the shock wave can be seen in condensation patterns: © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect MACH SPEEDS The vapor cone is explained by the Prandtl-Glauert singularity - a mathematical singularity that occurs in aerodynamics. The transonic wave front is rounded rather than pointed © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect MACH SPEEDS The Prandtl-Glauert singularity is the point at which a sudden drop in air pressure occurs, and is generally accepted as the cause of the visible condensation cloud that often surrounds an aircraft traveling at transonic speeds. One view of this phenomenon is that it exhibits the effect of compressibility and the so-called "N-wave". The N-wave is the time variant pressure profile seen by a static observer as a sonic compression wave passes. The overall three-dimensional shock wave is in the form of a cone with its apex at the supersonic aircraft. This wave follows the aircraft. The pressure profile of the wave is composed of a leading compression component (the initial upward stroke of the "N"), followed by a pressure descent forming a rarefaction of the air (the downward diagonal of the "N"), followed by a return to the normal ambient pressure (the final upward stroke of the "N"). © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect MACH SPEEDS These condensation clouds, also known as "shock-collars" or "shock eggs," are frequently seen during space-shuttle launches around 25 to 33 seconds after launch when the vehicle passes through the area of maximum dynamic air-pressure, or max Q. These effects are also visible in archival footage of some nuclear tests. The condensation marks the approximate location of the shock wave. © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect MACH SPEEDS Since heat does not leave the affected air mass, this change of pressure is adiabatic, with an associated change of temperature. In humid air, the drop in temperature in the most rarefied portion of the shock wave (close to the aircraft) can bring the air temperature below its dew point, at which moisture condenses to form a visible cloud of microscopic water droplets. Since the pressure effect of the wave is reduced by its expansion (the same pressure effect is spread over a larger radius), the vapor effect also has a limited radius. Such vapor can also be seen in low pressure regions during high–g subsonic maneuvers in humid conditions. © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect MACH SPEEDS © 2006 By Timothy K. Lund

Sound 14-5 The Doppler Effect MACH SPEEDS vs > v Supersonic Line tangent to spherical supersonic wave fronts vt vst sin  = v vs sin  = vt FYI: The ratio M = vs / v is called the Mach number. 1 M  Thus sin  = vst © 2006 By Timothy K. Lund