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Unformatted text preview: Chapter 10 Answer to problems 27, 29, 32, 43, 47, 48 , 62, 63 27..The period of oscillation of a springandmass system is 0.50 s and the amplitude is 5.0 cm. What is the magnitude of the acceleration at the point of maximum extension of the spring? Given: A=5.0cm=5.0x102 m, T=0.50s, Question: a at maximum extension ? At the maximum magnitude of the acceleration x=A the amplitude, v m = A, am= 2 A But = 2p / T a max = A (2p/T) 2 =(5.0x102 m)x[4 p 2 / (0.50s) 2 ] = 7.9 m/s 2 43.An ideal spring with a spring constant of 15 N/m is suspended vertically. A body of mass 0.60 kg is attached to the unstretched spring and released. (a) What is the extension of the spring when the speed is a maximum? (b) What is the maximum speed? Given: m=0.60kg, Question: a) x at v max ? b) v max ? It is at the equilibrium point that the speed is maximum and it is minimum when the Amplitude is reached. The forces are applied in the y direction Writing newtons law at equilibrium, SF y =0 a) SF y = kx mg = 0 x = mg / k =( 0.60kg x 9.80 m/s 2 ) /15N = 0.39 m b) v m = A = x v(k/m) = 0.39m v(15N/0.60kg) = 1.95 m/s 48..A body is suspended vertically from an ideal spring of spring constant 2.5 N/m. The spring is initially 48....
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This note was uploaded on 12/14/2011 for the course PHY 14434 taught by Professor Huang during the Spring '09 term at Wayne State University.
 Spring '09
 HUANG

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