Question

A 1 kg block is executing simple harmonic motion of amplitude 0.1 m on a smooth horizontal surface under the restoring force of a spring of spring constant 100 N/m. A block of mass 3 kg is gently placed on it at the instant it passes through the mean position. Assuming that the two blocks move together. Find the amplitude of the motion?

• A. $0.5m$
• B. $0.05m$
• C. $5m$
• D. $0.02m$

Answer 1

The correct option is B $0.05m$

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m_{combinedmass}}}=\frac{1}{2\pi }\sqrt{\frac{100}{4}}=\frac{2.5}{\pi }=0.8$ Hz
with 1 kg mass, $f_0=\frac{1}{2\pi }\sqrt{\frac{100}{1}}=\frac{5}{\pi }$ Hz
Further, from conservation of linear momentum (at mean position)
$\omega A=\frac{1}{4}{\omega }_0{A}_0$
or $fA=\frac{1}{4}{f}_0{A}_0$
or $A=\frac{f_0{A}_0}{4f}-\frac{\left(5/\pi \right)\left(0.1\right)}{\left(4\right)\left(2.5/\pi \right)}$
$=0.05m$