 PHY 1150
Chapter 14; Waves and Sound | ToC, Waves and Sound | Return to Calendar | D14.1 What is the wavelength of a 50 Hz wave when the wave speed is 340 m/s ?

v = f  = v / f  D14.2 Consider a 9.0 m cable with a mass of 5.4 kg. When the cable is struck, a pulse moves down the cable and back in 0.6 s. What is the tension in the cable?    T = v2 (m/L) = (30 m/s)2 (0.6 kg/m)
T = 540 N

D14.3 Sketch the progression of two pulses shown here. The "top" one is moving to the right and the "bottom" one is moving to the left. Each pulse moves at 10 cm/s.           D14.4

Sketch the progression of two pulses shown here. The "short" one is moving to the right and the "tall" one is moving to the left. Each pulse moves at 10 cm/s.      D14.5 The A string on a violin is tuned to a fundamental frequency of 440 Hz. List the overtones that may be present in the audible range of frequencies. f1 = 440 Hz 1 = (1/2) L n = (1/n) 1

or
fn = n f1

f1 = 440 Hz
f2 = 880 Hz
f3 = 1 320 Hz
f4 = 1 760 Hz
f5 = 2 200 Hz
f6 = 2 640 Hz
f7 = 3 060 Hz
f8 = 3 520 Hz
f9 = 3 960 Hz

etc
f45 20 000 Hz will be near or beyond the range of hearing.

D14.6 A guitar string 0.70 m long is tuned to play A at 440 Hz whe its full length vibrates. Where should a finger be placed in order to play C at 524 Hz?

The velocity of the wave on the string is given by

v = f  /2 = 0.70 m = 1.4 m
v = (440 Hz) (1.4 m)
v = 616 m/s  /2 = 0.588 m
That is, your finger should be placed so the string that vibrates is 0.588 m (or 0.59 m) long.

D14.7 A demonstration is carried out with the apparatus shown here. A mass is suspended on one end of a string (so the tension in the string is the weight of the mas, mg). The string is then run over a pulley and attached to a small-amplitude 60-Hz oscillator. As additional masses are placed on the end, standing waves appear. The distance between pulley and oscillator is 2.4 m. That amoun of string has a mass of 45 g. How much mass is supported when the standing wave -- with three "loops" -- shown in the diagram is produced?  /2 = 2.4 m / 3 = 0.80 m = 1.6 m
f = 60 Hz = 60 (
1/s)
v = f = (60/s) (1.6 m) = 96 m/s  T = v
2 (m/L) T = v
2 (m/L) = (96 m/s)2 (0.01875 kg/m) = 172.8 N
T = m g
m = T/g = (172.8 N ) / (9.8 m/s2)
m = 17.6 kg

D14.8 With the string and arrangement shown in the problem above, what hanging mass is necessary to produce a standing wave of five "loops" if the distance from pulley to oscillator remains at 2.4 m? /2 = 2.8 m/5 = 0.56 m = 1.12 m
f = 60 Hz = 60 (
1/s)
v = f = (60/s) (1.12 m) = 67.2 m/s  T = v
2 (m/L) T = v
2 (m/L) = (67.2 m/s)2 (0.01875 kg/m) = 84.67 N
T = m g m = 8.64 kg

D14.9 An organ pipe open at one end and closed at the other is 1.0 m long. What is its fundamental frequency? What is its first overtone frequency?  1 = 4.0 m
v = f   The next overtone is such an open pipe -- exactly like the resonance tubes we use in the lab -- has (3/4) of a wavelength.
( 3/4 ) 2 = 1.0 m 2 = 1.33 m D14.10 A ringing tuning fork is held above a tube in a Physics experiment. The tube is filled with water whose level can be altered easily, thus changing the length of the column of air below the tuning fork. As the water level is lowered, lengthening the column of air, an increase in loudness is noted when the water is 0.120, 0.360, and 0.600 m from the top of the tube. Assume the speed of sound in air is 345 m/s. What is the frequency of the tuning fork?

This one should be (very) familar!

( 1/2 ) = 0.600 m - 0.360 m = 0.360 m - 0.120 m = 0.240 m = 0.480 m
v = f  D14.11 What is the speed of sound in air on a very hot summer day when the temperature is 42o C?

v = [331 + (0.60)(42)] m/s

v = [331 + 25] m/s
v = 356 m/s

D14.12 A train whistle sounds at 500 Hz. What frequency is heard by a stationary observer when the train approaches at 25 m/s? When it moves away at 25 m/s?

What frequency is heard by a moving observer as she approaches the stationary train at 25 m/s? As she moves away from the stationary train at 25 m/s?

Take the speed of sound in air to be 340 m/s.  f ' = (500 Hz) (1.079) = 540 Hz f ' = (500 Hz) (0.932) = 466 Hz f ' = (500 Hz) (1.074) = 537 Hz f ' = (500 Hz) (0.926) = 463 Hz

D14.13 A stationary train engineer sounds his own engine's horn, which has a frequency of 550 Hz, as another train passes. He notices a beat frequency of 2 Hz both as the train approaches and as it departs. What is the speed of the moving train? Take the speed of sound in air to be 340 m/s. f = 550 Hz
v = 340 m/s
v
obs = 0
v
s = ? (we are looking for the velocity of the source).

A beat frequency of 2 Hz means the frequency heard from the moving train's horn must be different from 550 Hz by 2 Hz. Therefore, for the approaching train, the frequency heard must be 552 Hz, and for the train going away, the frequency heard must be 548 Hz.  340 = (1.0036)(340 - v
s) = 341.236 &endash; 1.0036 vs
1.0036 v
s = 1.236   340 = (0.9964)(340 + v
s) = 338.764 + 0.9964 vs
0.9964 v
s = 1.236 Remember, however, the "beat frequency of 2 Hz" is stated to only one significant figure. We are unjustified in keeping three significant figures in our answers.  