This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: h ic e8ac ice k# i b iC a aha 13 Problem 7 ( a ): If the frequency is f = 6 . 00 Hz = 6 . 00 oscillations per second, then the period is T = 1 /f = . 167 s. ( b ): The angular frequency is = 2 f = 37 . 7 / s. ( c ): We have = radicalbigg k m m = k 2 = 120 N / m (37 . 7 / s) 2 = 0 . 0844 kg . 13 Problem 26 ( a ): The total energy at this point is E = 1 2 mv 2 + 1 2 kx 2 = 1 2 (0 . 150 kg)(0 . 300 m / s) 2 + 1 2 (300 N / m)(0 . 0120 m) 2 = 0 . 0284 J . ( b ): The amplitude A is the maximum value of x , at which point all the energy is potential, so 1 2 kA 2 = E A = radicalbigg 2 E k = 0 . 0137 m . ( c ): When the object is at maximimum speed all the energy is kinetic, so 1 2 mv 2 max = E v max = 2 E m = 0 . 615 m / s . 13 Problem 36: The moment of inertia is I = 1 2 mR 2 = 4 . 84 10 7 kg m 2 . For a torsional pendulum we have T = 2 radicalbigg I = 4 2 I T 2 = 1 . 91 10 5 N m . 13 Problem 47: The period of oscillation is T = 1 . 36 s. We have T = 2 radicalBigg L g g = 4 2 L T 2 = 10 . 7 m / s 2 ....
View
Full
Document
This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
 Work

Click to edit the document details