Question

An air-track cart attached to a spring completes one oscillation every 2.4s. At t=0 the cart is released from rest at a distance of .10m from its equilibrium position. What is the position of the cart at 2.7s?

Answer 1

Position of cart with respect to equilibrium position is given by

$x = (0.1 m)cos \omega t$

  $= (0.1 m)cos (2 \pi/T) t$

  $= (0.1 m)cos (2 \pi(2.7)/2.4)$

  $= (0.1 m)cos (2.25 \pi)$

= (0.1 m)× 0.707

= 0.0707 m

= 7.07 cm

(Note that T is the time period and cart is performing simple harmonic motion . Therefore we have used standard equation of SHM)