Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
The displacement of a particle of mass 200 g from its mean position is given by \[y=0.05\text{ }\sin \text{ }(20\text{ }\pi t+1.6\pi )m\] (1) The period of motion is 0.1 s (2) The maximum acceleration is\[20{{\pi }^{2}}m{{s}^{-2}}\] (3) The energy of the particle is\[0.10{{\pi }^{2}}J\] (4) The energy of the particle is\[{{\pi }^{2}}J\]A) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: A
Solution :
The equation of SHM \[y=0.05\text{ }\sin (20\pi t+1.6\pi )\] ... (i) The standard equation of SHM \[y=A\sin (\omega t+\phi )\] ...(ii) On comparing Eqs. (i) and (ii), we get Amplitude, A = 0.05 m and angular velocity,\[\omega =20\text{ }\pi \text{ }rad/\text{ }s\] (i) Time period, \[T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }=0.1\text{ }s\] (ii) Maximum acceleration, \[{{a}_{\max }}={{\omega }^{2}}A\] \[={{(20\pi )}^{2}}\times 0.05\] \[=20{{\pi }^{2}}\,m/{{s}^{2}}\] (iii) Energy, \[E=\frac{1}{2}m{{\omega }^{2}}A\] \[E=\frac{1}{2}\times \frac{200}{1000}\times {{(20\pi )}^{2}}\times (0.05)\] \[=0.10{{\pi }^{2}}J\] Hence, option is correct.
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