Question

A mass $m$, which is attached to a spring with constant $k$, oscillates on a horizontal table, with amplitude $A$. At an instant when the spring is stretched by $\sqrt{3}A/2$, a second mass $m$ is dropped vertically onto the original mass and immediately sticks to it. What is the amplitude of the resulting motion?

• A. $\frac{\sqrt{3}}{2}A$
• B. $\sqrt{\frac{7}{8}}A$
• C. $\sqrt{\frac{13}{16}}A$
• D. $\sqrt{\frac{2}{3}}A$

Answer 1

The correct option is B $\sqrt{\frac{7}{8}}A$

Formula for the velocity $v$, of a particle/mass executing SHM with amplitude $A$, and at displacement $x$ is,
$v=\omega \sqrt{A^2-x^2}$
now at $x=\frac{A\sqrt{3}}{2}$, velocity will be, $\frac{A\omega }{2}$, at this point another mass $m$ attached to the mass vertically preserving linear momentum. that causes for the whole mass move with a new velocity $\frac{\omega A}{4}$ at $\frac{A\sqrt{3}}{2}$ and with a different angular velocity $\omega /\sqrt{2}$ as we know $\omega =\sqrt{\frac{k}{m}}$, mass doubles in the second situation.
Substituting new values of velocity and angular frequency in the equation of velocity at displacement $\frac{A\sqrt{3}}{2}$, new amplitude turns out to be $A\sqrt{\frac{7}{8}}$