A particle starts executing simple harmonic motion (SHM) of amplitude ‘a’ and total energy E. At any instant,
Question:

A particle starts executing simple harmonic motion (SHM) of amplitude ‘a’ and total energy E. At any instant, 

its kinetic energy is $\frac{3 \mathrm{E}}{4}$ then displacement ‘ $y$ ‘ is given by :

  1. y = a 

  2. $y=\frac{a}{\sqrt{2}}$

  3. $y=\frac{a \sqrt{3}}{2}$

  4. $y=\frac{a}{2}$


Correct Option: , 4

Solution:

$\mathrm{E}=\frac{1}{2} \mathrm{Ka}^{2}$

$\frac{3 \mathrm{E}}{4}=\frac{1}{2} \mathrm{~K}\left(\mathrm{a}^{2}-\mathrm{y}^{2}\right)$

$\frac{3}{4} \times \frac{1}{2} \mathrm{Ka}^{2}=\frac{1}{2} \mathrm{~K}\left(\mathrm{a}^{2}-\mathrm{y}^{2}\right)$

$y^{2}=a^{2}-\frac{3 a^{2}}{4}$

$y=\frac{a}{2}$

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