Question

A body is performing simple harmonic with an amplitude of $10cm$. The velocity of the body was tripled by air Jet when it is at $5cm$ from its mean position. The new amplitude of vibration is $\sqrt{x}$ $cm$. The value of $x$ is ______

Answer 1

Answer: 700

$A=10cm$
$\therefore$ Total Energy $=\frac{1}{2}K{A}^2$
By energy conservation we can final $v$ at $x=5$
$\frac{1}{2}K(10{)}^2=\frac{1}{2}K(5{)}^2+\frac{1}{2}m{v}^2$
$V=\sqrt{\frac{75K}{m}}$
Now, velocity is tripled through external mean so the amplitude of SHM will charge and so the total energy, (but potential) energy at this moment will remain same)
$\therefore \frac{1}{2}K(5{)}^2+\frac{1}{2}m{\left(3\sqrt{\frac{75K}{m}}\right)}^2=\frac{1}{2}K{A}^2$
$\Rightarrow 25K+675K=K{A}^2$
$\therefore A=\sqrt{700}$
$\therefore x=700$