Asked by Ayodeji

A mass of 0.5kg is attached to one end of a helical spring and produced an extension amplitudeof 2.5cm.The mass is now set into vertical oscillation of 10mm.Determine (a)the period of oscillation(b)the maximum kinetic energy of the mass(C)the potential energy of the spring when the mass is 8mm below the center of oscillation (g=10ms-2)
Answered by Anonymous
F = k x
.5 * 9.81 = k * 1.5 * 10^-2
k = 3.27* 10^2 = 327 Newtons/meter

{{{{to find period ----you might know f = (1/2pi)sqrt (k/m) but anyway:
if x = A sin wt where w = 2 pi f
v = Aw cos w t
a = -Aw*2 sin wt = -w^2 x but F = m a
so
- k x = - m w^2 x
w^2 = k/m
w = sqrt(k/m) = 2 pi f }}}
so
f = (1/2 pi) sqrt (k/m)
f = (1/2pi) sqrt (327 / .5) = (1/6.28) sqrt(654) = 4.07 Hz
T = 1/4.07 = 0.246 seconds (part a)
I think you can get the maximum potential energy from my note below. That is the max kinetic energy as well. However you can also get (1/2) m v^2 at x = 0 from my derivation above
potential energy of spring = (1/2) k x^2
Answered by James stacy
Who be rat abeg comot for road make person hear well e be like say you don dey smoke small small
Answered by James stacy
Forgive my rudeness I wall only messing around.
Answer
(Correction not wall I meant Was.) Sure thanks for understanding I appreciate.
Answered by Bot
I apologize if my response was not clear or understandable. Is there anything specific that you did not understand or need clarification on? I am here to assist you. And I assure you that I am an AI language model and do not smoke.
Answered by Bot
No problem at all! If you have any further questions or need help, feel free to ask.
Answered by Bot
You're welcome! If you have any questions, don't hesitate to ask.