A 0.480-kg object attached to a spring with a force constant of 8.00 N/m vibrate

Question

A 0.480-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 10.2 cm. (Assume the position of the object is at the origin at

t = 0.)

(a) Calculate the maximum value of its speed.
cm/s

(b) Calculate the maximum value of its acceleration.
cm/s2

(c) Calculate the value of its speed when the object is 6.20 cm from the equilibrium position.
cm/s

(d) Calculate the value of its acceleration when the object is 6.20 cm from the equilibrium position.
cm/s2

(e) Calculate the time interval required for the object to move from x = 0 to x = 2.20 cm.
s

Explanation / Answer

k=8
m=0.48
A=0.102 m
let equation of motion be:

x(t)=A*sin(w*t+theta)

where w=sqrt(k/m)=4.08 rad/s

then x(t)=0.102*sin(4.08*t+theta)

now at t=0, x=0
so theta=0

a)v(t)=A*w*cos(w*t)
maximum value=A*w=0.416 m/s
b)acceleration maximum=A*w^2=1.697 m/s^2

c)0.062=0.102*sin(4.08*t)

t=9.174 seconds

then speed=A*w*cos(w*t)=0.33 m/s

d)acceleration=-A*w^2*sin(w*t)=1.031 m/s^2

e)x=0.022
t=3.05 seconds

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