Q152 / 200 · Manipal (MET) 2012 · Physics

For a particle executing simple harmonic motion, the kinetic energy $K$ is given by, $K=K_0 \cos ^2 \omega t$. The maximum value of potential energy is

  • Correct answer
    A
    $K_0$
  • B
    zero
  • C
    $K_0 / 2$
  • D
    not obtainable

Solution

$K_{\max }=K_0=$ total energy As total energy remains conserved in SHM, hence when $U$ is maximum in $\mathrm{SHM}, K=0$, i.e., $E$ iş also equal to $U_{\max }$, i.e., $U_{\max }=E=K_0$.

Hence, the correct answer is option: 'A'.

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