Question

An object performing S.H.M. with mass of 0.5 kg, force constant 10N/m and amplitude 3cm, What is the speed at x = 2 cm?
(Ans : V = 0.1 m/s )

Answer 1

we know, kinetic energy of object performing S.H.M at x distance from mean position is given by $K.E=\frac{1}{2}\kappa(A^2-x^2)$
actually, maximum speed can be found at mean position. so, we have to take formula of kinetic energy of object at mean position.

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

$\frac{1}{2}mv^2=\frac{1}{2}\kappa(A^2-x^2)$

$mv^2=\kappa(A^2-x^2)$

given, m = 0.5 kg
$\kappa=10N/m$
$A=3cm=3\times10^{-2}m$

0.5 × v² = 10 × (0.03² - 0.02²)

0.5 × v² = 10 × (0.0009 - 0.0004)

0.5 × v² = 10 × 0.0005

0.5 × v² = 0.005

v² = 0.01

v = 0.1 m/s