Question

A particle starts SHM from the mean position $x=0$. Its amplitude is $a$ and total energy is $E$. When the particle is at a position $5\text{cm}$ from the mean position, its kinetic energy is $\frac{3E}{4}$. The value of $a$ (in $\text{cm}$) is
[Assume, The potential energy of the particle is zero at the mean position]

Answer 1

Kinetic energy of the particle at a position $x$ from the mean position is given by $K=\frac{1}{2}k[a^2-x^2]$
From the data given in the question,
$\frac{1}{2}k[a^2-x^2]=\frac{3E}{4}$
But, $E=\frac{1}{2}k{a}^2$
$\therefore$ we get, $a^2-x^2=\frac{3a^2}{4}$
$x^2=\frac{a^2}{4}$ or $x=\pm \frac{a}{2}$
Given that, $x=5\text{cm}$
$\Rightarrow |a|=2x=2\times 5=10\text{cm}$