Question

A I kg block is executing simple harmonic motion of amplitude 0. I 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 frequency and the amplitude of the motion.

Answer 1

Amplitude $=01m$

Total mass $=3+1=4Kg$ (When both blocks moving together)

$\therefore T=2\pi \sqrt{\frac{M}{K}}=2\pi \sqrt{\frac{4}{100}}=\frac{2\pi }{5}sec$

$\therefore$ frequency $=\frac{5}{2\pi }Hz$

Again at mean position, let $1Kg$ block has velocity $v$

$K.E.=\frac{1}{2}m{v}^2=\frac{1}{2}m{x}^2$

where $x\to$ amplitude $=0.1m$

$\therefore \frac{1}{2}\times 1\times v^2=\frac{1}{2}\times 100\times (0.1{)}^2$

$v=1m/sec$ .....(1)

After the $3Kg$ block is gently placed on $1Kg$, then let $1Kg+3Kg=4Kg$ block and the spring be one system.

$\therefore$ initial momentum $=$ final momentum

$1timesv=4\times v^{\prime }$

$v^{\prime }=\frac{1}{4}m/s$ [from 1]

Now, the two blocks have velocity $\frac{1}{4}m/s$ at its mean position

$K{E}_{mass}=\frac{1}{2}m.v^{\prime 2}=\frac{1}{2}\times 4\times {\left(\frac{1}{4}\right)}^2=\frac{1}{2}\times \frac{1}{4}$

When the block are going to the extreme position, there will be only $PE$

$\therefore PE=\frac{1}{2}K\delta =\frac{1}{2}\times \frac{1}{4}$

where, $\delta \to$ new amplitude

$\frac{1}{4}=100{\delta }^2\to \delta \sqrt{\frac{1}{400}}=0.05m$

Thus, amplitude $=5cm$