Question

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

• A. Amplitude of the motion becomes $5\text{cm}$
• B. The oscillation frequency becomes $\frac{5}{2\pi }\text{Hz}$
• C. Amplitude of the motion becomes $10\text{cm}$
• D. The oscillation frequency becomes $\frac{5}{\pi }\text{Hz}$

Answer 1

Answer: (A), (B)

The correct options are
A Amplitude of the motion becomes $5\text{cm}$
B The oscillation frequency becomes $\frac{5}{2\pi }\text{Hz}$

Given $A=0.1\text{m}$
$k=100\text{N/m}$
$m_1=1\text{kg}$
$m_2=3\text{kg}$

At the mean position, let $1\text{kg}$ block has velocity $v$
Energy of system is conserved
$\frac{1}{2}K{A}^2=\frac{1}{2}{m}_1{v}_1^2$
$\Rightarrow v_1=\frac{K{A}^2}{m_1}=\frac{100\times {0.1}^2}{1}$
$\Rightarrow v_1=1\text{m/s}$
At the moment $m_1$ is at mean position,
applying law of conservation of momentum,
$m_1{v}_1=(m_1+m_2)v$
$\Rightarrow 1=(m_1+m_2)v$
$\Rightarrow v=\frac{1}{(m_1+m_2)}=\frac{1}{4}=0.25\text{m/s}$

New amplitude as $3\text{kg}$ falls on $1\text{kg}$ is
$\frac{1}{2}{\text{KA'}}^2=\frac{1}{2}(m_1+m_2)v^2\Rightarrow A^{\prime 2}=\frac{(0.25{)}^2\times 4}{100}$
$\Rightarrow A^{\prime }=5\text{cm}$

Frequency of the combined mass performing oscillation is given by
$f=\frac{1}{2\pi }\sqrt{\frac{k}{m_1+m_2}}$
$f=\frac{1}{2\pi }\sqrt{\frac{100}{4}}$
$f=\frac{5}{2\pi }\text{Hz}$
Thus, option (a) and (b) are correct.