A) 0.4 J
B) 0.5 J
C) 3 J
D) 0.3 J
Correct Answer: D
Solution :
[d] \[V=\frac{dy}{dt}=5\times 4cos\,\left( 4\left( \frac{T}{4} \right)+\frac{\pi }{3} \right)\] \[V=20cos\,\left( T+\frac{\pi }{3} \right)\] As \[y=5\,\sin \,\left( 4t+\frac{\pi }{3} \right)\] Compare with \[y=A\,\sin \,\left( \omega t+\frac{\pi }{3} \right)\] \[\omega =4\Rightarrow T=\frac{2\pi }{4}=\frac{\pi }{2}\] So that \[V=20\text{ }cos\left( \frac{\pi }{2}+\frac{\pi }{3} \right)\] \[=-20\sin \frac{\pi }{3}=-10\sqrt{3}\] \[KE=\frac{1}{2}m{{u}^{2}}=\frac{1}{2}{{(10\sqrt{3})}^{2}}\times 2\times {{10}^{-3}}=0.3J\]
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