15 MECHANICAL WAVES 15 1 1 T v f IDENTIFY f is the time for one complete vibration SET UP The frequency of the note one octave higher is 1568 Hz EXECUTE a 0 439 m 1 28 ms T v f 344 m s 784 Hz 1 f v f 344 m s 1568 Hz 0 219 m b EVALUATE When f is doubled is halved IDENTIFY The distance between adjacent dots is frequency 20 0 Hz and the short wavelength sound has the highest frequency 20 0 kHz SET UP For sound in air v f 15 2 The long wavelength sound has the lowest EXECUTE a Red dots Blue dots 344 m s 3 20 0 10 Hz v 344 m s v f 344 m s 17 2 m 20 0 Hz 0 0172 m 1 72 cm b In each case the separation easily can be measured with a meterstick c Red dots 74 0 m v f 1480 m s 3 1480 m s 20 0 Hz 20 0 10 Hz Blue dots 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 IDENTIFY v SET UP 1 0 h T The crest to crest distance is f 3600 s 3 800 10 m 220 m s 3600 s EXECUTE v v 800 km 800 km h 1 0 h EVALUATE Since the wave speed is very high the wave strikes with very little warning IDENTIFY f SET UP 1 0 mm 0 0010 m 1500 m s 0 0010 m v v f EXECUTE 1 5 10 Hz 6 EVALUATE The frequency is much higher than the upper range of human hearing IDENTIFY We want to relate the wavelength and frequency for various waves SET UP For waves v f 15 3 15 4 15 5 Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher 15 1 15 2 Chapter 15 EXECUTE a v 344 m s For f 20 000 Hz 1 7 cm For f 20 Hz v f 344 m s 20 000 Hz v f 344 m s 17 m 20 Hz The range of wavelengths is 1 7 cm to 17 m b v c 3 00 10 m s For 8 700 nm f 4 3 10 Hz For 14 400 nm c 3 00 10 m s 700 10 m 8 9 7 5 10 Hz 14 The range of frequencies for visible light is 4 3 10 Hz 14 to 3 00 10 m s 400 10 m 8 9 c f 7 5 10 Hz 14 c v 344 m s d v 1480 m s v f 3 344 m s 23 10 Hz 1480 m s 23 10 Hz 3 v f 1 5 cm 6 4 cm 15 6 EVALUATE For a given v a larger f corresponds to smaller increases IDENTIFY The fisherman observes the amplitude wavelength and period of the waves SET UP The time from the highest displacement to lowest displacement is highest displacement to lowest displacement is 2A The distance between wave crests is of the waves is T 2 T v f For the same f increases when v The distance from and the speed 15 7 and k is defined by Eq 15 5 The general form of the EXECUTE a T 2 2 5 s 5 0 s 6 0 m v 6 0 m 1 2 m s 5 0 s T A v f x A y v f 1 25 0 Hz 0 320 m direction x cos2 A t 0 0 62 m 2 0 31 m f 1 2 2 rad 0 320 m 19 6 rad m 0 0700 m 8 00 m s 0 320 m b c The amplitude becomes 0 15 m but the wavelength period and wave speed are unchanged EVALUATE The wavelength period and wave speed are independent of the amplitude of the wave IDENTIFY Use Eq 15 1 to calculate v wave function is given by Eq 15 8 which is the equation for the transverse displacement A SET UP 8 00 m s v f EXECUTE a so T 1 25 0 Hz 0 0400 s k b For a wave traveling in the t T Eq 15 8 y x t 0 x t T so y A at At cos2 Substitute in numerical values y x t x Or c From part b Plug in y y d In part c 0 150 s y A means cos 2 t T x n 0 2 4 cos 0 1 2 So y A when 2 n x t T t T x n t T n x 0 0400 s 0 0400 s 0 360 m 0 320 m 4 n t For 0 1150 s 0 0700 m cos 2 0 320 m 0 0400 s and 0 0700 m cos 2 0 360 m 0 320 m 0 150 s 0 0400 s 4 95 cm 0 0700 m cos 2 4 875 rad t 157 rad s t 0 0700 m cos 2 0 320 m 0 0400 s 1 0 0700 m cos 19 6 m before the instant in part c 1 n 2 n 2 y 0 360 m t x x 1 for 0 0495 m 0 150 s or or 1 125 y x t 0 t t n x t This equation describes the wave specified in the problem Copyright 2012 Pearson Education Inc All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced in any form or by any means without permission in writing from the publisher Mechanical Waves 15 3 15 8 15 9 5 n 0 150 s t 0 1550 s t for it to travel from the first occurrence of y A after the instant in part c Thus the elapsed time For is 0 1550 s 0 1500 s 0 0050 s EVALUATE Part d says y A at 0 115 s and next at 0 155 s the difference between these two times is 0 040 s which is the period At T upward It takes 4 0 0100 s reasonable IDENTIFY Compare SET UP The comparison gives EXECUTE a 6 50 mm b 28 0 cm 1 c 0 360 m y A so our answer of 0 0050 s is y x t given in the problem to the general form of Eq 15 4 the particle at 0 y and traveling 6 50 mm v 0 0360 s 4 95 cm 28 0 cm 27 8 Hz x to T and 1 f is …