A 0.450 kg object attached to a springwith a force constant of 8.00 N/m vibrates

Question

A 0.450 kg object attached to a springwith a force constant of 8.00 N/m vibrates in simple harmonicmotion with an amplitude of 11.5 cm. (a) Calculate the maximum value (magnitude) ofits speed and acceleration.
cm/s
cm/s2
(b) Calculate the speed and acceleration when the object is6.50 cm from the equilibriumposition.
cm/s
cm/s2
(c) Calculate the time interval required for the object to movefrom x = 0 to x = 3.50 cm.
s
(a) Calculate the maximum value (magnitude) ofits speed and acceleration.
cm/s
cm/s2
(b) Calculate the speed and acceleration when the object is6.50 cm from the equilibriumposition.
cm/s
cm/s2
(c) Calculate the time interval required for the object to movefrom x = 0 to x = 3.50 cm.
s

Explanation / Answer

A 0.450 kg object attached to a springwith a force constant of 8.00 N/m vibrates in simple harmonicmotion with an amplitude of 11.5 cm.
x(t)= Acos(t)

v(t) =x'(t)= -Asin(t)
a(t) =x''(t)= -A2cos(t)
=(k/m) = (8.00/0.450)
=4.22 rad/s
(a) Calculate the maximumvalue (magnitude) of its speed and acceleration.
cm/s v(t) =x'(t)= -Asin(t) '; vmax= A = 11.5 x 4.22= 48.5cm/s
a(t) =x''(t)= -A2cos(t); amax= A2= 11.5 x (4.22)2 =205 cm/s2
(b) Calculate the speed andacceleration when the object is 6.50 cm from the equilibriumposition.
cm/s x(t)=Acos(t) ; compute time at x=6.50cm 6.50=11.5cos(4.22t) 6.50/11.5 =cos(4.22t) t= (1/4.22) arccos(6.50/11.5) (just remember it must be inradians) t= (1/4.22) (0.9700) t= 0.230 sec v(t)= -Asin(t) ; v(0.230)=-11.5 x 4.22 sin(4.22 x 0.230) v(0.23)= 40.0 cm/s Similarly Similarly a(t) =-A2cos(t);   a(0.23) = -11.5(4.22)2cos(4.22 x 0.23) =116m/s2 (c) Calculate the timeinterval required for the object to move from x = 0 tox = 3.50 cm.
t=t2-t1 since t1=0 t=t2=t again t= (1/4.22) arccos(3.5/11.5)=0.299 s

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