An object with mass 2.0kg is attached to a spring with spring stiffness constant 240N/m and is executing simple harmonic motion. When the object is 2.2×10−2m from its equilibrium position, it is moving with a speed of 0.65m/s . Calculate the amplitude of the motion.

Answer 1

x=2.2•10⁻² m
v=0.65 m/s
m=2 kg
k =240 N/m =>
ω=sqrt(k/m) = sqrt(240/2) =11 rad/s
x=Asin ωt
v= ω Acos ωt
tan ωt = xω /v=2.2•10⁻²•11/0.65 =0.372
ωt =20.4°
sin ωt=0.349
x=Asin ωt
A=x/sin ωt =2.2•10⁻²/0.349=6.3•10⁻² m