Question

An object performing S.H.M. with mass of $0.5kg$, force constant $10N/m$ and amplitude $3cm$ . What is the speed at x=2 ?

Answer 1

We know, kinetic energy of object performing S.H.M at x distance form

mean position is given by K.E =$\frac{1}{2}k(A^2-x^2)$

actually, maximum speed can be found at mean position.

so we have to taken formula of kinetic energy of object at mean position.

but we know. kinetic energy K.E = $\frac{1}{2}m{v}^2$

$\frac{1}{2}m{v}^2=\frac{1}{2}k(A^2-x^2)$

$m{v}^2=k(A^2-x^2)$

given, m = 0.5 kg

k = 10 N/m

A = 3cm = $3\times {10}^{-2}m$

$0.5\times v^2=10\times ({0.03}^2-{0.02}^2)$

$0.5\times v^2=10\times (0.0009-0.0004)$

$0.5\times v^2=10\times 0.0005$

$0.5\times v^2=0.005$

$v^2=0.01$

v = 0.1 m/s