A) 0.4 J
B) 0.5 J
C) 3 J
D) 0.3 J
Correct Answer: D
Solution :
The displacement of particle, executing SHM is \[y=5\sin \left( 4t+\frac{\pi }{3} \right)\] ?.(i) Velocity of particle \[\left( \frac{dy}{dt} \right)=\frac{5d}{dt}\,\sin \,\left( 4t+\frac{\pi }{3} \right)\] \[=5\cos \left( 4t+\frac{\pi }{3} \right).4\] \[=20\,\cos \left( 4t+\frac{\pi }{3} \right)\] Velocity at \[t=\left( \frac{T}{4} \right)\] \[{{\left( \frac{dy}{dt} \right)}_{t=\frac{T}{4}}}=20\cos \left( 4\times \frac{T}{4}+\frac{\pi }{3} \right)\] or \[u=20\cos \left( T+\frac{\pi }{3} \right)\] ??(ii) Comparing the given equation with standard equation of SHM, given by \[y=a\,\sin (\omega t+\phi )\] We get \[\omega =4\] As \[\omega =\frac{2\pi }{T}\] \[\Rightarrow \] \[T=\frac{2\pi }{\omega }\] or \[T=\frac{2\pi }{4}\] or \[T=\left( \frac{\pi }{2} \right)\] Now, putting value of T in Eq. (ii), we get \[u=20\,\cos \,\left( \frac{\pi }{2}+\frac{\pi }{3} \right)\] \[=-20\sin \frac{\pi }{3}\] \[=-20\times \frac{\sqrt{3}}{2}\] \[=-10\times \sqrt{3}\] The kinetic energy of particle, \[KE=\frac{1}{2}m{{u}^{2}}\] \[\because \] \[m=2g=2\times {{10}^{-3}}g\] \[=\frac{1}{2}\times 2\times {{10}^{-3}}\times {{(-10\sqrt{3})}^{2}}\] \[={{10}^{-3}}\times 100\times 3\] \[=3\times {{10}^{-1}}\] \[KE=0.3J\]
You need to login to perform this action.
You will be redirected in
3 sec
