Question

A 25 kg object attached to an ideal spring with a force constant spring constant) of 15 N/m oscillates on a horizontal, frictionless track. At timet0.00 s the cart is released from restat position x3 cm from the equilibrium position le) Express the displacement as a function of time using a cosine function Note: If you do not know how to write it using the editor, please submit your answer as part of your work-up document Arwer

Answer 1

spring mass system is

d. m E-ko dt2
  $(\frac{d^{2}x}{dt^{2}} = a)$

$\frac{d^{2}x}{dt^{2}} +\frac{k}{m}x=0$

$\frac{d^{2}x}{dt^{2}} +\frac{15}{2.5}x=0$

$\frac{d^{2}x}{dt^{2}} +6x=0$

auxiliary eqn is

$m^{2}=-6$

$m= \pm i\sqrt{6}$ or   $m= 0\pm i\sqrt{6}$

then solution is

$x= e^{\alpha t }(C_{1}cos\beta t+C_{2}sin\beta t)$

$x= e^{0.t }(C_{1}cos\beta t+C_{2}sin\beta t)$

$x= (C_{1}cos\sqrt{6} t+C_{2}sin\sqrt{6}t)$----------------------------------(1)

now initial condition

  

( b/c x at t=0 is  constant so v=0 at t=0 )

Differentiate  the eqn(1)

$v=\frac{dx}{dt}= \sqrt{6}(-C_{1}Sin\sqrt{6} t+C_{2}Cos\sqrt{6}t)$

using intial condtion

v(t=0)=0 m/s

$0= \sqrt{6}C_{2}$

$C_{2}=0$--------------------------(2)

and x(t=0)=8cm

from eqn (1)

-------------------------(3)

put the eqn(2) and eqn (3) in eqn (1)

$x=0.08Cos\sqrt{6}t$