Question

A body mass $1$ kg is suspended from a weightless spring having force constant $600$N/m.Another body of mass $0.5kg$ moving vertically upwards hits the suspended body with a velocity of $3.0m/s$ and get embedded in it.Find the frequency of oscillations and amplitude of motion.

• A. $10/\pi Hz,\frac{5\sqrt{37}}{6}cm$
• B. $5/\pi Hz,\frac{5\sqrt{37}}{6}cm$
• C. $10/\pi Hz,\frac{5\sqrt{37}}{7}cm$
• D. $10/\pi Hz,\frac{7\sqrt{37}}{6}cm$

Answer 1

The correct option is A $10/\pi Hz,\frac{5\sqrt{37}}{6}cm$

we know that

here $m=1.0+0.5=1.5kg$

$k=600N/m$

and

$w=20$

Hz

now in a collision momentum get conserved so

and finally bullet is embedded

so $\left(0.5\right)\left(3\right)=\left(1.5\right)\left(v\right)$

so $v=1m/s$

this system will do SHM

and we know that

here A is the amplitude of the SHM and x is the present position

actually the mean position of the SHM is

$x_1=\frac{mg}{k}$

so $x_1=1.5\times \frac{10}{600}$

$x_1=\frac{1}{40}$

but right now the block is at $x_2=1\times 10/600=1/60$

difference in the position or distance from the mean position is $x_1-x_2$

that is $x=1/120$

so

$A=$ $cm$