Question
The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is $$10\,{s^{ - 1}}.$$ At, $$t = 0$$ the displacement is $$5\,m.$$ The initial phase is $$\frac{\pi }{4}.$$ What is the maximum acceleration ?
A.
$$500\,m/{s^2}$$
B.
$$500\sqrt 2 \,m/{s^2}$$
C.
$$750\,m/{s^2}$$
D.
$$750\sqrt 2 \,m/{s^2}$$
Answer :
$$500\sqrt 2 \,m/{s^2}$$
Solution :
$${\text{Given}}\,{\text{that,}}\,\,\frac{{{A_{\max }}}}{{{v_{\max }}}} = 10\,\,{\text{i}}{\text{.e}}{\text{.,}}\,\,\omega = 10\,{s^{ - 1}}$$
Displacement is given by
$$\eqalign{
& x = a\sin \left( {\omega t + \frac{\pi }{4}} \right) \cr
& {\text{at}}\,\,t = 0,x = 5;5 = a\sin {45^ \circ } \Rightarrow a = 5\sqrt 2 \cr
& {A_{\max }} = a{\omega ^2} = 500\sqrt 2 \,m/{s^2} \cr} $$