Class Management I Help Simple Harmonic Motion (CH 16) Begin Date: 11/26/2016 12

Question

Class Management I Help Simple Harmonic Motion (CH 16) Begin Date: 11/26/2016 12:00:00 AM -Due Date: 12/2/2016 11:59:00 PM End Date: 12/2/2016 i 59: a4%) Problem 4: A mass m 15 kg hangs at the end of a vertical spring whose top end is fixed to the ceiling. The spring has spring constant k 85 Nlm and negligible mass. At time t* 0 the mass is released from rest at a distance d 0.25 m below its equilibrium height and undergoes simple harmonic motion with its position given as a function of time by Acos(oor- The positive y-axis points upward. otheexpertta.c A 17% Part (a) Find the angular frequency of oscillation, in radians per second. 17% Part (b) Determine the value of the coefficient A, in meters. A 17% Part (e) Determine the value of in radians. 17% Part (d) Enter an expression for the velocity along they axis as a function of time, in terms ofA.ao, and 17% Part (e) What is the velocity, in meters per second, at time 0.15 s? 17% Part (n What is the magnitude of the mass s maximum acceleration, in meters per second squared? Grade Summary Potential 100 7 8 9 tano sin0 cos() Attempts remaining. asino acos0 per cotano atano acotano sinh0 1 2 3 cosh() oDegrees Radians

Explanation / Answer

Here ,

m = 1.5 Kg

k = 85 N/m

d = 0.25 m

part A)

angular frequency = sqrt(k/m)

angular frequency = sqrt(85/1.5)

angular frequency = 7.53 rad/s

part B)

A = d = 0.25 m

part C)

for the value of phi

as the block is at the maximum negative position as t = 0

phi = pi radian = 3.14 radian

part d)

for the velocity

vy = -A * sin(w * t)

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please post other parts in a seperate post

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