Question

A block attached to a horizontal spring of force constant 75N/m undergoes SHM with an amplitude of 0.15m. If the speed of the mass is 1.7 m/s when the displacement is 0.12m from the equilibrium position, what is the mass of the block?

Answer 1

x = A sin w t for SHM where w = angular frequency

sin w t = x / A = .12 / .15 = .8 where w t is the angle in degrees

w t = 53.1 deg (w itself is in rad / sec)

since v = A w cos w t

then v / x = w cos w t / sin w t = w / tan w t

w = v / x * tan 53.1 = 1.7 / .12 * tan 53.1 = 18.9 /sec

Also for SHM w = (k / m)^1/2

m = k / w^2 = 75 / 18.9^2 = .21 kg