A) \[12\,{{\omega }^{2}}\]
B) \[36\,{{\omega }^{2}}\]
C) \[144\,{{\omega }^{2}}\]
D) \[\sqrt{192}\,{{\omega }^{2}}\]
Correct Answer: B
Solution :
\[x=12\sin \omega \,t-16\,{{\sin }^{3}}\omega \,t=4[3\sin \omega \,t-4{{\sin }^{3}}\omega \,t]\] \[=4[\sin 3\omega \,t]\] (By using \[\sin 3\theta =3\sin \theta -4{{\sin }^{3}}\theta )\] \[\therefore \] maximum acceleration \[{{A}_{\max }}={{(3\omega )}^{2}}\times 4=36{{\omega }^{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
