JEE MAIN - Physics (2017 (Offline) - No. 1)
A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of
equilibrium. The kinetic energy – time graph of the particle will look like:
_en_1_1.png)
_en_1_2.png)
_en_1_3.png)
_en_1_4.png)
Explanation
For a particle executing SHM
At mean position; t = 0, $$\omega $$t = 0, y = 0, V = Vmax = a$$\omega $$
$$ \therefore $$ K.E = K.Emax = $${1 \over 2}m{\omega ^2}{a^2}$$
At extreme position : t = $${T \over 4}$$ , $$\omega $$t = $${\pi \over 2}$$ , y = A, V = Vmin = 0
$$ \therefore $$ K.E = K.Emin = 0
Hence graph (A) correctly represents kinetic energy - time graph.
At mean position; t = 0, $$\omega $$t = 0, y = 0, V = Vmax = a$$\omega $$
$$ \therefore $$ K.E = K.Emax = $${1 \over 2}m{\omega ^2}{a^2}$$
At extreme position : t = $${T \over 4}$$ , $$\omega $$t = $${\pi \over 2}$$ , y = A, V = Vmin = 0
$$ \therefore $$ K.E = K.Emin = 0
Hence graph (A) correctly represents kinetic energy - time graph.
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