Question

the position of a 0.30 kg object attached to a spring is described by x=(0.25 m)cos(0.4π t).
find:

A.) the amplitude of the motion
B.) the spring constant
C.)the position at t= 0.30s
D.) the object speed at t=0.30s

Answer 1

Answer on question #64336, Physics / Mechanics — Relativity

Question the position of a 0.30 kg object attached to a spring is described by x=(0.25 m)cos(0.4 t). find:

A.) the amplitude of the motion

B.) the spring constant

C.)the position at t= 0.30s

D.) the object speed at t=0.30s

Solution So the equation of motion is

$x(t)=A\cos wt=0.25\cos(0.4\pi t)$

A. The amplitude is

$A=0.25\,m$

B. Spring constant

$k=mw^{2}=0.3\cdot(0.4\pi)^{2}\approx 0.473\,N/m$

C. Position at t = 0.3:

$x(0.3)=0.25\cos(0.4\pi\cdot 0.3)\approx 0.232\,m$

D.Speed is

$v(t)=\dot{x}(t)=-Aw\sin wt=-0.1\pi\sin(0.4\pi t)$

$v(0.3)=-0.1\pi\sin(0.4\pi\cdot 0.3)\approx-0.116\,m/s$

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