A) 9.80 m/s and \[9.03\times {{10}^{2}}m/{{s}^{2}}\]
B) 8.90 m/s and \[8.21\times {{10}^{2}}m/{{s}^{2}}\]
C) 6.82 m/s and \[7.62\times {{10}^{2}}m/{{s}^{2}}\]
D) 5.65 m/s and \[5.32\times {{10}^{2}}m/{{s}^{2}}\]
Correct Answer: D
Solution :
Key Idea: When particle passes through its equilibrium position, then the velocity is maximum and acceleration is maximum at extreme position. The velocity of a particle in SHM is given by \[u=\omega =\sqrt{{{a}^{2}}-{{y}^{2}}}\] where \[\omega \] is angular velocity, a the amplitude and y the displacement. When the particle passes through its equilibrium position, then the velocity is maximum. \[{{u}_{\max }}=a\,\omega \] ... (i) Similarly maximum acceleration at maximum displacement \[{{\alpha }_{\max }}={{\omega }^{2}}a\] Given, \[a=0.06\,m,\,\,\omega =2\,\pi \,f=2\pi \times 15\] Putting these values in Eqs. (i) and (ii), we get \[{{u}_{\max }}=0.06\times 2\pi \times 15\] \[=0.06\times 2\times 3.14\times 15\] \[{{\alpha }_{\max }}=5.32\times {{10}^{2}}m/{{s}^{2}}\]
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