Question

A block of mass $10\text{kg}$ attached to a horizontal spring with force constant $250\text{N/m}$ is moving with simple harmonic motion having amplitude $10\text{m}$. At the instant when the block passes through its equilibrium position, a lump of putty with mass $5\text{kg}$ is dropped vertically on the block from a very small height and sticks to it. Find the new amplitude and period?

• A. $8.1\text{m},1.54\text{s}$
• B. $10\text{m},1.54\text{s}$
• C. $8.1\text{m},1.25\text{s}$
• D. $12.2\text{m},1.54\text{s}$

Answer 1

The correct option is A $8.1\text{m},1.54\text{s}$

Given,
mass of block $\left(M\right)=10\text{kg}$
mass of pulley $\left(m\right)=5\text{kg}$
Amplitude $\left(A_1\right)=10\text{m}$
Force constant $\left(K\right)=250\text{N/m}$

Time period of oscillation as the putty drops on to the mass
$\left(T_2\right)=2\pi \sqrt{\frac{M+m}{K}}=2\times 3.14\times \sqrt{\frac{15}{250}}=1.54\text{s}$

Let $v_1$ be the speed of block at equilibrium position,
$v_2$ be the speed of the combined mass,
Momentum of the system in horizontal direction is conserved, then
$(M+m)v_2=M{v}_1$
$\therefore v_2=\left(\frac{M}{M+m}\right)v_1...\left(2\right)$
Now, let $A_2$ be the amplitude of system afterwards, then
$\therefore A_2{\omega }_2=\left(\frac{M}{M+m}\right)A_1{\omega }_1$
$A_2=\left(\frac{M}{M+m}\right)A_1\frac{{\omega }_1}{{\omega }_2}$
$A_2=\left(\frac{M}{M+m}\right)A_1\sqrt{\frac{M+m}{M}}$
$A_2=A_1\sqrt{\frac{M}{M+m}}$
$=10\times \sqrt{\frac{10}{15}}=10\sqrt{\frac{2}{3}}=8.1\text{m}$