Question

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

Answer 1

$A =10\, cm$
$\therefore$ Total Energy $=\frac{1}{2} KA ^{2}$
By energy conservation we can final $v$ at $x=5$
$\frac{1}{2} K (10)^{2}=\frac{1}{2} K (5)^{2}+\frac{1}{2} mv ^{2}$
$ V =\sqrt{\frac{75 \,K }{ 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)
$\therefore \frac{1}{2} K (5)^{2}+\frac{1}{2} m \left(3 \sqrt{\frac{75\, K }{ m }}\right)^{2}=\frac{1}{2} KA ^{2}$
$\Rightarrow 25 \,K +675\, K = KA ^{2}$
$\therefore A =\sqrt{700}$
$\therefore x =700$