- Home
- Standard 11
- Physics
A block of mass $m$ attached to massless spring is performing oscillatory motion of amplitude $'A'$ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $fA.$ The value of $f$ is
$\frac{1}{2}$
$\frac{1}{\sqrt{2}}$
$1$
$\sqrt{2}$
Solution
At equilibrium position
$V _{0}=\omega_{0} A =\sqrt{\frac{ K }{ m }} A….(i)$
$V =\omega A '=\sqrt{\frac{ K }{\frac{ m }{2}}} A '…(ii)$
$\therefore \quad A'= \sqrt{2} A$
$P_{i}=P_{f}$
$mV_0=\frac m{2} V$
$m(A \omega)=\frac{m}{2}\left(A^{\prime} \omega^{\prime}\right)$
$m A \sqrt{\frac{k}{m}}=\frac{m}{2} A^{\prime} \sqrt{\frac{k}{m / 2}}$
$m^{2} A^{2}\left[\frac{K}{m}\right]=\frac{m^{2}}{42}\left(A^{\prime}\right)^{2} \frac{k}{m} \times v$
$2 A^{2}=\left(A^{\prime}\right)^{2} \Rightarrow A^{\prime}=\sqrt{2}$
