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Q.
The ratio of displacement to amplitude, when the kinetic energy of a body executing SHM is thrice the potential energy is,
NTA AbhyasNTA Abhyas 2020Oscillations
Solution:
Kinetic energy in SHM,
$K=\frac{1}{2} m \omega^2\left(a^2-x^2\right)$
Potential energy in SHM,
$U=\frac{1}{2} m \omega^2 x^2$
Given that, $K=3U$
$\Rightarrow \quad \frac{1}{2} m \omega^2\left(a^2-x^2\right)=3 \cdot \frac{1}{2} m \omega^2 x^2$
$a^{2}-x^{2}=3x^{2}$
$a^{2}=4x^{2}$
$a=2x$
$2x=a$
$\frac{x}{a}=\frac{1}{2}$