JEE MAIN - Physics (2021 - 27th July Morning Shift - No. 13)
A particle starts executing simple harmonic motion (SHM) of amplitude 'a' and total energy E. At any instant, its kinetic energy is $${{3E} \over 4}$$ then its displacement 'y' is given by :
y = a
$$y = {a \over {\sqrt 2 }}$$
$$y = {{a\sqrt 3 } \over 2}$$
$$y = {a \over 2}$$
Explanation
$$E = {1 \over 2}K{a^2}$$
$${{3E} \over 4} = {1 \over 2}K({a^2} - {y^2})$$
$$ \Rightarrow $$ $${3 \over 4} \times {1 \over 2}K{a^2} = {1 \over 2}K({a^2} - {y^2})$$
$$ \Rightarrow $$ $${y^2} = {a^2} - {{3{a^2}} \over 4}$$
$$ \Rightarrow $$ $$y = {a \over 2}$$
$${{3E} \over 4} = {1 \over 2}K({a^2} - {y^2})$$
$$ \Rightarrow $$ $${3 \over 4} \times {1 \over 2}K{a^2} = {1 \over 2}K({a^2} - {y^2})$$
$$ \Rightarrow $$ $${y^2} = {a^2} - {{3{a^2}} \over 4}$$
$$ \Rightarrow $$ $$y = {a \over 2}$$
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