| [a] velocity will be one-half its maximum velocity |
| [b] displacement will be one-half its amplitude |
| [c] acceleration will be nearly \[86\,%\] of its maximum acceleration |
| [d] \[KE\,\,=\,\,PE\] |
A) a, b
B) b, c
C) a, c
D) c, d
Correct Answer: C
Solution :
\[V=A\omega \,\,cos\text{ }\omega t\] \[V={{V}_{\max }}\,cos\text{ }\frac{2\pi }{T}\times \frac{T}{6}\] \[V={{V}_{\max }}\,.\,\,\frac{1}{2}\] \[V\,\,=\,\,\frac{{{V}_{\max }}}{2}\] \[\because \,\,a=-A{{\omega }^{2}}\,\sin \,\,\omega t\] \[a=-{{a}_{\max }}\,\sin \,\,\frac{2\pi }{T}\,\,.\,\,\frac{T}{6}\] \[a=-{{a}_{\max }}\,.\,\frac{\sqrt{3}}{2}\] \[a\,\,=\,\,-0.866\,{{a}_{max}}\]
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