A) Amplitude of SHM. is \[\sqrt{\frac{2E}{A}}\]
B) Maximum velocity of the particle during S.H.M. is \[\sqrt{EB}\]
C) Time period of motion is \[2\pi \sqrt{\frac{B}{A}}\]
D) Displacement of the particle is proportional to the velocity of the particle.
Correct Answer: C
Solution :
Amplitude is obtained for \[v=0\] |
\[\therefore \]Amplitude \[=\sqrt{\frac{E}{A}}\] |
Maximum velocity is obtained for \[x=0\] |
\[{{v}_{\max }}=\sqrt{\frac{\operatorname{E}}{B}}\] |
\[{{v}_{\max }}=\operatorname{amplitude}\times \,\omega \]\[\Rightarrow \]\[\omega =\sqrt{\frac{A}{B}}\Rightarrow T=2\pi \sqrt{\frac{B}{A}}\] |
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