A) T/6
B) T/12
C) T/8
D) T/24
Correct Answer: B
Solution :
\[KE=\,3PE\] \[KE=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})\] \[PE=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[({{A}^{2}}-{{x}^{2}})\,=3{{x}^{2}}\] \[x=A/2\] At t = 0, y = 0 so \[\phi =0\] \[x=A\sin \,\omega t\,=A\sin \,\left( \frac{2\pi }{T}\times t \right)\] \[x=A/2\,\,\,for\,\,t=T/12\] \[({{A}^{2}}-{{x}^{2}})\,=3{{x}^{2}}\] \[x=A/2\] At \[t=0,\,\,y=0,\] so \[\phi =0\] \[x=A\sin \,\omega t\,=A\sin \left( \frac{2\pi }{T}\times t \right)\] \[x=A/2\,\] For \[t=T/12\]
You need to login to perform this action.
You will be redirected in
3 sec
