Question

A body of mass $1kg$ is executing simple harmonic motion. Its displacement ${}^{\prime }{y}^{\prime }$ cm at ${}^{\prime }{t}^{\prime }$ seconds is given by $y=6sin100t+\pi /4$. It maximum kinetic energy is:

Answer 1

We have,

$y=6sin\left(100t\right)+\pi /4$

$V=\frac{dy}{dt}=\left[6sin\right(100t+\pi /4\left)\right]$

$V=600cos(100t+\pi /4)$

$KE=\frac{M{V}^2}{2}=\frac{M}{2}\left(600{)}^{2}co{s}^2\right(100t+\pi /4)$

As $0⩽co{s}^2(100t+\pi /4)⩽1,$

$K{E}_{max}=\frac{m}{2}(600{)}^2=\frac{1}{2}(600{)}^2$

$=360000J=360KJ$