question_answer 58)

                  A mass of 2 kg is attached to the spring of spring constant \[50\,\,N{{m}^{-1}}\]. The block is pulled to a distance of 5 cm from its equilibrium position at \[x=0\] on a horizontal frictionless surface from rest at \[t=0\]. Write the expression for its displacement at anytime \[t\].

Answer:

                  Here, amplitude, \[r=5\,cm\,=5\times {{10}^{-2}}\,m\].                 \[\omega =\,\sqrt{\frac{k}{M}}\,=\,\sqrt{\frac{50}{2}}\,=\,5\,\,rad\,\,{{s}^{-1}}\]                 Using x \[x=r\,\sin \,\omega t,\] we get                                \[x=(5\times {{10}^{-2}}\,\sin \,5t)\,m\]