A) \[\frac{64}{9}m\]
B) \[\frac{1024}{9}m\]
C) \[\frac{3}{32}m\]
D) \[\frac{32}{9}m\]
Correct Answer: D
Solution :
Key Idea: When particle passes through equilibrium position, velocity is maximum and acceleration is maximum at maximum displacements The velocity of a particle in SHM changes with displacement y of the particle; When \[y=0\], velocity is maximum given by \[{{v}_{\max }}=a\,\omega \] ... (i) But acceleration is maximum, when y = a \[\therefore \] \[{{\alpha }_{\max }}={{\omega }^{2}}\alpha \] ?. (ii) where a is amplitude. Squaring Eq. (i) and dividing by Eq. (ii), we get \[\frac{{{v}^{2}}_{\max }}{{{\alpha }_{\max }}}=a\]. Given, \[{{v}_{\max }}=16\,m/s,\,{{\alpha }_{\max }}=24\,m/{{s}^{2}}\] \[\therefore \] \[a=\frac{{{(16)}^{2}}}{24}=\frac{32}{3}\,m\]
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