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Please answer all parts of the question. The multiple choice options are given in the images. (10%) Problem 6: A 085-kg mass oscillates according to the equation x(t)-0.72 cos(5.5t), where the position x(t) is measured in meters 25% Part (a) What is the period, in seconds, of this mass? Grade Summary Deductions Potential 0% 100% cos) sinO cotan asi atan()a Submissions Attempts remaining: 10 (10%) per attempt) detailed view 4 5 6 acos) sinh0 acotan() cosh tanh cotanh) Degrees Radians END DEL CLEAR Submit Hint I give up! Hints: 5% deduction per hint. Hints remaining: 2 Feedback: 0%-deduction per feedback. 25% Part (b) At what point during the cycle is the mass moving at it\'s maximum speed? ? ? 25% Part (c) what is the maximum acceleration of the mass, in meters per square second? 25% Part (d) At what point in the cycle will it reach it\'s maximum acceleration? Solution mass=m=0.85 kg position x=0.72*cos(5.5*t) part a: comparing with standard SHM equation x(t)=A*cos(w*t) where A=amplitude w=angular frequency here A=0.72 m w=5.5 rad/s then time period=2*pi/w =1.1424 seconds part b: velocity=dx/dt=-0.72*5.5*sin(5.5*t) so maximum velocity will occur when sin(5.5*t)=1==>cos(5.5*t)=0 ==>x=0 hence maximum speed at the equilibrium point in the middle of the cycle part c: acceleration=dv/dt=-0.72*5.5*5.5*cos(5.5*t) maximum acceleration=0.72*5.5*5.5=21.78 m/s^2 part d: maximum acceleration will be when cos(5.5*t)=1==>x is maximum hence at maximum displacement of the cycle. .

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velocity=dx/dt=-0.72*5.5*sin(5.5*t)
so maximum velocity will occur when sin(5.5*t)=1==>cos(5.5*t)=0
==>x=0
hence maximum speed at the equilibrium point in the middle of the cycle
part c:
acceleration=dv/dt=-0.72*5.5*5.5*cos(5.5*t)
maximum acceleration=0.72*5.5*5.5=21.78 m/s^2
part d:
maximum acceleration will be when cos(5.5*t)=1==>x is maximum
hence at maximum displacement of the cycle.

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