ssm_ch15

# Ssm_ch15 - Chapter 15 Student Solutions Manual 3(a The amplitude is half the range of the displacement or xm = 1.0 mm(b The maximum speed vm is

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Chapter 15 – Student Solutions Manual 3. (a) The amplitude is half the range of the displacement, or x m = 1.0 mm. (b) The maximum speed v m is related to the amplitude x m by v m = ω x m , where is the angular frequency. Since = 2 π f , where f is the frequency, ( )( ) 3 = 2 = 2 120 Hz 1.0 10 m = 0.75 m/s. mm vf x ππ × (c) The maximum acceleration is () ( ) ( ) ( ) 2 2 23 = = 2 = 2 120 Hz 1.0 10 m = 5.7 10 m/s . m ax f x ωπ π ×× 2 2 7. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/ T = 1/(0.500 s) = 2.00 Hz. (c) The angular frequency is = 2 π f = 2 π (2.00 Hz) = 12.6 rad/s. (d) The angular frequency is related to the spring constant k and the mass m by = km . We solve for k and obtain k = m 2 = (0.500 kg)(12.6 rad/s) 2 = 79.0 N/m. (e) Let x m be the amplitude. The maximum speed is v m = x m = (12.6 rad/s)(0.350 m) = 4.40 m/s. (f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by F m = kx m = (79.0 N/m)(0.350 m) = 27.6 N. 9. The magnitude of the maximum acceleration is given by a m = 2 x m , where is the angular frequency and x m is the amplitude. (a) The angular frequency for which the maximum acceleration is g is given by = gx m / , and the corresponding frequency is given by 2 6 11 9 . 8 m / s 498 Hz. 2 2 2 1.0 10 m m g f x == = = × (b) For frequencies greater than 498 Hz, the acceleration exceeds g for some part of the motion.

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17. The maximum force that can be exerted by the surface must be less than μ s F N or else the block will not follow the surface in its motion. Here, μ s is the coefficient of static friction and F N is the normal force exerted by the surface on the block. Since the block does not accelerate vertically, we know that F N = mg , where m is the mass of the block. If the block follows the table and moves in simple harmonic motion, the magnitude of the maximum force exerted on it is given by F = ma m = m ω 2 x m = m (2 π f ) 2 x m , where a m is the magnitude of the maximum acceleration, is the angular frequency, and f is the frequency. The relationship = 2 π f was used to obtain the last form. We substitute F = m (2 π f ) 2 x m and F N = mg into F < μ s F N to obtain m (2 π f ) 2 x m < μ s mg . The largest amplitude for which the block does not slip is = 2 = 0.50 9.8 22 . 0 0 031 2 2 2 x g f m s ππ b g b gc h b g m/s Hz × = .. m A larger amplitude requires a larger force at the end points of the motion. The surface cannot supply the larger force and the block slips. 19. (a) Let = 2 2 1 x At T cos π F H G I K J be the coordinate as a function of time for particle 1 and = 2 2 + 6 2 x T cos F H G I K J be the coordinate as a function of time for particle 2. Here T is the period. Note that since the range of the motion is A , the amplitudes are both A /2. The arguments of the cosine functions are in radians. Particle 1 is at one end of its path ( x 1 = A /2) when t = 0. Particle 2 is at A /2 when 2 π t / T + π /6 = 0 or t = – T /12. That is, particle 1 lags particle 2 by one- twelfth a period. We want the coordinates of the particles 0.50 s later; that is, at t = 0.50 s, 1 20 . 5 0
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## This note was uploaded on 04/09/2008 for the course PHYS 131 taught by Professor Drty during the Spring '08 term at Lafayette.

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Ssm_ch15 - Chapter 15 Student Solutions Manual 3(a The amplitude is half the range of the displacement or xm = 1.0 mm(b The maximum speed vm is

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