Question

A block weighing 17 N oscillates at one end of a vertical spring for which k = 120 N/m; the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched 0.24 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?

Answer 1

If v=0, the block is at its extreme (end) position =>
x= x(max)=0.24 m = A (amplitude)

The equation of the oscillation is
x=Asinωt
v=Aωcosωt
a= - Aω²sinωt
F= - m Aω²sinωt = - mω²x = - kx,
x=A => F= kA = 120∙0.24= 28.8 N.
ω= √(k/m)= √(kg/W)=
=√(120∙9.8/17) = 8.32 rad/s
T=2π/ ω=2π/8.32=0.76 s.
KE(max) =PE(max) =kA²/2 = 120∙0.24²/2 = 3.456 J.

Answer 2

No the amplitude is not 0.24 m read the question carefully, u will notice that 0.24 m is beyond its relaxed length (the length when no object is attached


so the correct amplitude will be

0.24m-(17N/120N/m) = 0.0983m