PHYS Homework # 10 (Mendes, Fall 2015) due in class on Nov 20. 1) Exercise 15.4, p. 501, University Physics by Young & Freedman

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PHYS 98 - Homework # 10 (Mendes, Fall 015) due in class on Nov 0 1) Exercise 154, p 501, Universit Phsics b Young & Freedman IDENTIFY: f v SET UP: 10 mm 00010 m v 1500m/s 6 EXECUTE: f 15 10 Hz 00010 m EVALUATE: The frequenc is much higher than the upper range of human hearing ) Exercise 158, p 501, Universit Phsics b Young & Freedman IDENTIFY: Compare ( x, t) given in the problem to the general form of Eq (154) f 1/ T and v f SET UP: The comparison gives EXECUTE: (a) 650 mm (b) 80 cm (c) f 1 78 Hz 00360 s (d) v (080 m)(78 Hz) 778 m/s A 650 mm, 80 cm and T 00360 s (e) Since there is a minus sign in front of the tt / term, the wave is traveling in the x -direction EVALUATE: The speed of propagation does not depend on the amplitude of the wave 3) Exercise 1514, p 50, Universit Phsics b Young & Freedman IDENTIFY: v and a are given b Eqs (159) and (1510) SET UP: The sign of v determines the direction of motion of a particle on the string If v 0 and a 0 the speed of the particle is increasing If v 0, the particle is speeding up if v and a have the same sign and slowing down if the have opposite signs EVALUATE: (a) The graphs are given in Figure 1514 (b) (i) v A sin(0) 0 and the particle is instantaneousl at rest particle is speeding up a A cos(0) A and the (ii) v Asin( /4) A /, and the particle is moving up a A cos( /4) A/, and the particle is slowing down ( v and a have opposite sign) (iii) v Asin( /) A and the particle is moving up instantaneousl not accelerating a Acos( /) 0 and the particle is (iv) v Asin(3 /4) A /, and the particle is moving up a Acos(3 /4) A/, and the particle is speeding up 1/1

(v) v A sin( ) 0 and the particle is instantaneousl at rest particle is speeding up a Acos( ) A and the (vi) v Asin(5 /4) A / and the particle is moving down a Acos(5 /4) A/ and the particle is slowing down ( v and a have opposite sign) (vii) v Asin(3 /) A and the particle is moving down particle is instantaneousl not accelerating (viii) v Asin(7 /4) A /, and the particle is moving down a Acos(3 /) 0 and the a Acos(7 /4) A/ and the particle is speeding up ( v and a have the same sign) EVALUATE: At t 0 the wave is represented b Figure 1510a in the textbook: point (i) in the problem corresponds to the origin, and points (ii) (viii) correspond to the points in the figure labeled 1 7 Our results agree with what is shown in the figure Figure 1514 4) Exercise 1516, p 50, Universit Phsics b Young & Freedman IDENTIFY: The frequenc and wavelength determine the wave speed and the wave speed depends on the tension SET UP: v F ml / v f 010 kg EXECUTE: F v ( f ) ( 400 Hz0750 m ) 43 N 50 m EVALUATE: If the frequenc is held fixed, increasing the tension will increase the wavelength 5) Exercise 15, p 50, Universit Phsics b Young & Freedman IDENTIFY: Appl Eq (155) SET UP: f ml / /1

EXECUTE: (a) P 1 av F A Pav 080 m figures 3 1 30010 kg (50 N)( (100 Hz)) (16 10 3 m) 03 W or 0 W to two (b) P av is proportional to A, so halving the amplitude quarters the average power, to 0056 W EVALUATE: The average power is also proportional to the square of the frequenc 6) Exercise 1530, p 503, Universit Phsics b Young & Freedman IDENTIFY: The distance the wave shape travels in time t is vt The wave pulse reflects at the end of the string, at point O SET UP: The reflected pulse is inverted when O is a fixed end and is not inverted when O is a free end EXECUTE: (a) The wave form for the given times, respectivel, is shown in Figure 1530a (b) The wave form for the given times, respectivel, is shown in Figure 1530b EVALUATE: For the fixed end the result of the reflection is an inverted pulse traveling to the left and for the free end the result is an upright pulse traveling to the left Figure 1530 7) Exercise 1534, p 503, Universit Phsics b Young & Freedman IDENTIFY: Appl the principle of superposition SET UP: The net displacement is the algebraic sum of the displacements due to each pulse EXECUTE: The shape of the string at each specified time is shown in Figure 1534 EVALUATE: The pulses interfere when the overlap but resume their original shape after the have completel passed through each other 3/1

Figure 1534 8) Exercise 1540, p 504, Universit Phsics b Young & Freedman L IDENTIFY: For a string fixed at both ends, n n and v fn n L SET UP: For the fundamental, n 1 For the second overtone, n 3 For the fourth harmonic, n 4 (48 0 m/s) EXECUTE: (a) 1 L 300 m f1 v 160 Hz L (150 m) (b) 3 1 /3 1 00 m f 3 f 1 480 Hz (c) 4 1 /4 0 75 m f3 4 f 1 640 Hz EVALUATE: As n increases, decreases and f increases 9) Exercise 1548, p 504, Universit Phsics b Young & Freedman 4/1

IDENTIFY: ( x, t) ( A sin kx)sin t v / t SW a / t SET UP: vmax ( ASW sin kx ) amax ( ASW sin kx ) EXECUTE: (a) (i) x is a node, and there is no motion (ii) x is an antinode, and 4 vmax A( f ) fa, amax ( f ) vmax 4 f A (iii) cos 1 and this factor multiplies 4 the results of (ii), so vmax fa, amax f A (b) The amplitude is Asin kx, or (i) 0, (ii) A, (iii) A / (c) The time between the extremes of the motion is the same for an point on the string (although the period of the zero motion at a node might be considered indeterminate) and is 1/ f EVALUATE: An point in a standing wave moves in SHM All points move with the same frequenc but have different amplitude 10) Exercise 157, Universit Phsics b Young & Freedman IDENTIFY: The time between positions 1 and 5 is equal to T / v f The velocit of points on the string is given b Eq (159) 60 s SET UP: Four flashes occur from position 1 to position 5, so the elapsed time is 4 0 048 s 5000 The figure in the problem shows that L 0500 m At point P the amplitude of the standing wave is 15 cm EXECUTE: (a) T / 0048 s and T 0096 s f 1/ T 104 Hz 0 500 m (b) The fundamental standing wave has nodes at each end and no nodes in between This standing wave has one additional node This is the 1st overtone and nd harmonic (c) v f (10 4 Hz)(0500 m) 50 m/s (d) In position 1, point P is at its maximum displacement and its speed is zero In position 3, point P is passing through its equilibrium position and its speed is vmax A fa (10 4 Hz)(0015 m) 0980 m/s (e) v F FL m and FL (100 N)(0500 m) m 185 g v (50 m/s) EVALUATE: The standing wave is produced b traveling waves moving in opposite directions Each point on the string moves in SHM, and the amplitude of this motion varies with position along the string Note: equal credit for each problem 5/1