Q192 / 200 · Manipal (MET) 2013 · 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

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

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

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