Question

In the figure shown, find the maximum amplitude of the system, so that there is no slipping between any of the blocks.

• A. $\frac{2}{7}\text{m}$
• B. $\frac{3}{4}\text{m}$
• C. $\frac{4}{9}\text{m}$
• D. $\frac{10}{3}\text{m}$

Answer 1

The correct option is C $\frac{4}{9}\text{m}$

We know that, $\omega =\sqrt{\frac{k}{\Sigma m}}=\sqrt{\frac{54}{6}}=3\text{rad/s}$
Maximum friction between blocks $A$ and $B$
$\left(f_1\right)=0.6\times 1\times 10=6\text{N}$

Maximum friction between blocks $B$ and $C$
$\left(f_2\right)=0.4\times 3\times 10=12\text{N}$

Maximum acceleration of block $A$ $\left(a_1\right)=\frac{f_1}{m_1}=6{\text{m/s}}^2$
Maximum acceleration of block $B$ $\left(a_2\right)=\frac{f_2}{m_2+m_1}=\frac{12}{3}=4{\text{m/s}}^2$

So, maximum acceleration of the whole system, so that there is no slipping between any of blocks is $4{\text{m/s}}^2$

Since, $a_{max}=\text{A}{\omega }^2$, we get
$\text{A}=\frac{a_{max}}{{\omega }^2}=\frac{4}{(3{)}^2}=\frac{4}{9}\text{m}$

Thus, option (c) is the correct answer.