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-Secure https://session.masteringphysics.com/myct/itemView?offset=next&assignmentProblemlD-84407173; Homework 04 (Chapter 15) Problem 15.12 Problem 15.12 Part A An object in simple harmonic motion has amplitude 7.8 cm and frequency 0.50 Hz . At0 s it has its most negative position. Write the function z(t) that describes the object's position. 2(t)-(3.9cm) cos((0.50 rad/s)t + rad) z(t)-(7.8cm) cos((1.00 rad/s)t- rad) z(t)-(7.8cm) cos((1.00 rad/s)t-/2 rad) (t)(7.8em) cos(0.50 rad/s)t) Submit My Answers Give Up Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

Here ,

amplitude , A = 7.8 cm

w = 2pi * f

w = 2pi * 0.50

w = pi rad/s

Now , as the object is initially at negative extreme position

x(t) = A * cos(wt + phi)

x(t) = 7.8 * cos(pi * t rad/s - pi)

the correct option is 2) 7.8 * cos(pi * t rad/s - pi)

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