YF_ISM_15

# YF_ISM_15 - MECHANICAL WAVES 15 15.1. IDENTIFY: v = f . T =...

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15-1 M ECHANICAL W AVES 15.1. IDENTIFY: vf λ = . 1/ Tf = is the time for one complete vibration. SET UP: The frequency of the note one octave higher is 1568 Hz. EXECUTE: (a) 344 m/s 0.439 m 784 Hz v f == = . 1 1.28 ms T f . (b) 344 m/s 0.219 m 1568 Hz v f = . EVALUATE: When f is doubled, is halved. 15.2. IDENTIFY: The distance between adjacent dots is . = . The long-wavelength sound has the lowest frequency, 20.0 Hz, and the short-wavelength sound has the highest frequency, 20.0 kHz. SET UP: For sound in air, 344 m/s v = . EXECUTE: (a) Red dots: 344 m/s 17.2 m 20.0 Hz v f = . Blue dots: 3 344 m/s 0.0172 m 1.72 cm 20.0 10 Hz = × . (b) In each case the separation easily can be measured with a meterstick. (c) Red dots: 1480 m/s 74.0 m 20.0 Hz v f = . Blue dots: 3 1480 m/s 0.0740 m 7.40 cm = × . In each case the separation easily can be measured with a meterstick, although for the red dots a long tape measure would be more convenient. EVALUATE: Larger wavelengths correspond to smaller frequencies. When the wave speed increases, for a given frequency, the wavelength increases. 15.3. IDENTIFY: / T λλ . SET UP: 1.0 h 3600 s = . The crest to crest distance is . EXECUTE: 3 800 10 m 220 m/s 3600 s v × . 800 km 800 km/h 1.0 h v . EVALUATE: Since the wave speed is very high, the wave strikes with very little warning. 15.4. IDENTIFY: fv = SET UP: 1.0 mm 0.0010 m = EXECUTE: 6 1500 m s 1.5 10 Hz 0.0010 m v f = × EVALUATE: The frequency is much higher than the upper range of human hearing. 15.5. IDENTIFY: = . = . SET UP: 9 1 nm 10 m = EXECUTE: (a) 400 nm = : 8 14 9 3.00 10 m/s 7.50 10 Hz 400 10 m c f × × . 15 1/ 1.33 10 s ==× . 700 nm = : 8 14 9 4.29 10 Hz 700 10 m f × × × . 15 2.33 10 s T . The frequencies of visible light lie between 14 × and 14 × . The periods lie between 15 1.33 10 s × and 15 2.33 10 s × . (b) T is very short and cannot be measured with a stopwatch. EVALUATE: Longer wavelength corresponds to smaller frequency and larger period. 15

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15-2 Chapter 15 15.6. IDENTIFY: Compare (,) yxt given in the problem to the general form of Eq.(15.4). 1/ fT = and vf λ = SET UP: The comparison gives 6.50 mm A = , 28.0 cm = and 0.0360 s T = . EXECUTE: (a) 6.50 mm (b) 28.0 cm (c) 1 27.8 Hz 0.0360 s f == (d) (0.280 m)(27.8 Hz) 7.78 m s v (e) Since there is a minus sign in front of the / tT term, the wave is traveling in the -direction x + . EVALUATE: The speed of propagation does not depend on the amplitude of the wave. 15.7. IDENTIFY: Use Eq.(15.1) to calculate v . Tf = and k is defined by Eq.(15.5). The general form of the wave function is given by Eq.(15.8), which is the equation for the transverse displacement. SET UP: 8.00 m/s, v = 0.0700 m, A = 0.320 m = EXECUTE: (a) = so / (8.00 m/s)/(0.320 m) 25.0 Hz fv = 1/ 1/ 25.0 Hz 0.0400 s = 2 / 2 rad/0.320 m 19.6 rad/m k πλ π = (b) For a wave traveling in the -direction, x (, ) c o s2(/ / ) yx t A x =+ (Eq.(15.8).) At 0, x = (0, ) cos2 ( / ), ytA t T = so yA = at 0. t = This equation describes the wave specified in the problem. Substitute in numerical values: ( , ) (0.0700 m)cos(2 ( /0.320 m /0.0400 s)). x t Or, 1 (0.0700 m)cos((19.6 m ) (157 rad/s) ). x t (c) From part (b), (0.0700 m)cos(2 ( /0.320 m /0.0400 s)). yx t Plug in 0.360 m x = and 0.150 s: t = (0.0700 m)cos(2 (0.360 m/0.320 m 0.150 s/0.0400 s)) y (0.0700 m)cos[2 (4.875 rad)] 0.0495 m 4.95 cm y + = + (d) In part (c) 0.150 s. t = = means cos(2 ( / / )) 1 xt T += cos 1 θ = for 0, = 2, 4, ( 2) n ππ = or 0, 1, 2, n = So = when 2(/ /) ( T n or / / xx T n ( / ) (0.0400 s)( 0.360 m/0.320 m) (0.0400 s)( 1.125) nx n n =− = = For 4, n = 0.1150 s t = (before the instant in part (c)) For 5, n = 0.1550 s t = (the first occurrence of = after the instant in part (c)) Thus the elapsed time is 0.1550 s 0.1500 s 0.0050 s.
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## This note was uploaded on 03/14/2012 for the course MAE 162D taught by Professor Shaefer during the Spring '11 term at UCLA.

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YF_ISM_15 - MECHANICAL WAVES 15 15.1. IDENTIFY: v = f . T =...

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