Question

The given arrangement is released from rest when spring is in natural length. Maximum extension in spring during the motion is$x_0.a_1,a_2$ and $a_3$ are accelerations of the blocks. Make the correct option.

• A. $a_2-a_1=a_1-a_3$
• B. $x_0=\frac{4mg}{3k}$
• C. $v=\frac{x_0}{2}\sqrt{\frac{3k}{14m}}$
• D. Acceleration$a_{1}at\frac{x_0}{4}is\frac{3k{x}_0}{42m}$

Answer 1

Answer: (A), (C)

$v=\frac{x_0}{2}\sqrt{\frac{3k}{14m}}$

Step 1: Draw the required diagram and given data:

The given arrangement is released from rest when spring is in natural length.

Maximum extension in spring during the motion is $x_o$

Acceleration of the block connected to the spring and on the surface is $a_1$

Acceleration of the blocks $mand2m$ hanging through the pulley is $a_{2}and{a}_3$

Step 2: Calculate the acceleration:

Note that from the figure,

From constraint relations hanging masses C and D have acceleration of

Mass C has an acceleration of $a_2-a_1$in the downward direction

Mass D has an acceleration of $a_3-a_1$in the downward direction

Considering mass 2m is moving downward

$\begin{array}{rcl}{a}_2-a_1& =& -\left(a_3-a_1\right)\\ a_2-a_1& =& -a_3+a_1\\ a_2-a_1& =& a_1-a_3\end{array}$

Thus, option A is the correct answer.

Step 3: Calculating equivalent mass

Using formula, $\frac{T}{g}=\frac{2m_1{m}_2}{m_1+m_2}$

where

$$\begin{gathered} m_1=m \\ {m}_2=2m \end{gathered}$$

$\begin{array}{rcl}\frac{T}{g}& =& \frac{2\left(2m\right)m}{\left(2m+m\right)}\\ \frac{T}{g}& =& \frac{4m}{3}\end{array}$

Multiply both sides by $2$

$\begin{array}{rcl}\frac{2T}{g}& =& \frac{8m}{3}\end{array}$

This way we found the equivalent mass

Thus equivalent mass will be $\frac{8m}{3}$

Step 4: Calculating maximum extension:

Let the final extension be $x_o$

At maximum extension, the velocity of the system is zero. So energy stored in the spring is work done by gravity.

Now using conservation of energy

$\begin{array}{rcl}\left(\frac{1}{2}\right)k{x_0}^2& =& \frac{8mg}{3}{x}_o\\ x_0& =& \frac{16mg}{3k}\end{array}$

Thus, the maximum extension will be$\begin{array}{rcl}& & \frac{16mg}{3k}\end{array}$.

Step 5: Calculating velocity:

System will be in simple harmonic motion,

Total mass $2m+\frac{8m}{3}$

The angular frequency will be

$\omega =\sqrt{\frac{k}{2m+{\displaystyle \frac{8m}{3}}}}\dots \dots \dots \left(1\right)$

For SHM the maximum extension will be -

$maximumextension=2\times amplitude$

The amplitude of oscillation will be,

$\begin{array}{rcl}A& =& \frac{x_o}{2}............\left(2\right)\\ A& =& \frac{{\displaystyle \frac{16mg}{3k}}}{2}\\ & =& \frac{8mg}{3k}\end{array}$

The velocity will be given as

$v=A\omega$

From equations (1) and (2)

$\begin{array}{rcl}{V}_{x0/2}& =& V_{max}=\frac{x_0}{2}\omega \\ & =& \frac{x_0}{2}\sqrt{\frac{k}{2m+{\displaystyle \frac{8m}{3}}}}\\ & =& \frac{x_0}{2}\sqrt{\frac{3k}{14m}}\end{array}$

Hence option C is also correct.

Step 6: Calculating acceleration:

Acceleration will be calculated as

$a=-A{\omega }^2$

At $A=\frac{x_o}{4}$, the acceleration $a_1$will be

$\begin{array}{rcl}{a}_1& =& \frac{x_0}{4}\times {\omega }^2\\ & =& \frac{x_0}{4}\times \frac{3k}{14m}\\ & =& \frac{3k{x}_0}{56m}\end{array}$ From equation (1) substituting the value of angular velocity

Hence, Options A, and C are correct answers.