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NEET Sample Paper NEET Sample Test Paper-85

  • question_answer
    A particle of mass of 4 kg suspended from a spring of force constant \[800\,\,N\,\,{{m}^{-1}}\] executes simple harmonic oscillations. If the total energy of the oscillator is 4 J, the maximum acceleration \[(in\,\,m\,{{s}^{-2}})\] of the particle is

    A) 5                     

    B)        15

    C) 45                    

    D)        20

    Correct Answer: D

    Solution :

    [d] Here,\[m=4kg;\]\[k=800N{{m}^{-1}};\]\[E=4J\] In SHM, \[E=\frac{1}{2}k{{A}^{2}}\] \[\therefore \]      \[4=\frac{1}{2}\times 800\times {{A}^{2}}\] \[{{A}^{2}}=\frac{8}{800}=\frac{1}{100},\]\[A=0.1\,\,m\] Maximum acceleration,\[{{a}_{\max }}={{\omega }^{2}}A\] \[=\frac{k}{m}A\]    \[\left( \therefore \omega =\sqrt{\frac{k}{m}} \right)\] \[=\frac{800N{{m}^{-1}}}{4kg}\times 0.1\,\,m=20\,m{{s}^{-2}}\]


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