A block of mass is attached to a massless spring with spring constant . This block is connected to two other blocks of masses and using two massless pulleys and strings. [2019]

Question

A block of mass $2 M$ is attached to a massless spring with spring constant $k$ . This block is connected to two other blocks of masses $M$ and $2 M$ using two massless pulleys and strings. The acceleration of the blocks are $a_{1}$ , $a_{2}$ and $a_{3}$ as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x_{0}$ . Which of the following option(s) is/are correct [ $g$ is the acceleration due to gravity. Neglect friction] [2019]

Smart Achievers · Accepted Answer

(4)

According to constraint relation from figure,

$a_{1} = \frac{a_{2} + a_{3}}{2}$

$\Rightarrow a_{2} + a_{3} = 2 a_{1}$

$\Rightarrow a_{2} + a_{3} = a_{1} + a_{1}$

$\Rightarrow a_{1} - a_{3} = a_{2} - a_{1}$

$\Rightarrow Option (d) is correct$

Let $' x'$ be the extension of the spring at a certain instant,

$2 T - K x = 2 M a_{1}$

$2 M g - T = 2 M a_{3}$

$M g - T = M a_{2}$

On solving we get,

$a_{1} = \frac{4 g}{7} - \frac{3 k x}{14 M} = - \frac{3 K}{14 M} (x - \frac{8 M g}{3 K})$ ...(i)

Comparing it with $a = - ω^{2} (x - x_{0})$ ,

$∴$ $ω^{2} = \frac{3 k}{14 M}$ $∴$ $ω = \sqrt{\frac{3 k}{14 M}}$

and $T = \frac{4 M g}{7} + \frac{2 k x}{7}$ ...(ii)

For $a_{1} = 0$ (Maximum extension of spring) we have from (i),

$\frac{4 g}{7} - \frac{3 k x}{14 M} = 0$

$∴$ $4 g = \frac{3 k x}{2 M}$ $∴$ $x = \frac{8 M g}{3 k}$

$∴$ $x_{0} = 2 x = \frac{16 M g}{3 k}$

For $x = \frac{x_{0}}{4} = \frac{1}{4} (\frac{16 M g}{3 k}) = \frac{4 M g}{3 k}$

From eqn. (i), $a_{1} = \frac{4 g}{7} - \frac{3 k}{14 M} \times \frac{4 M g}{3 k} = \frac{2 g}{7}$

At $x = \frac{x_{0}}{2}$ , particle is at mean position and its velocity $= A ω$

$= \frac{x_{0}}{2} \sqrt{\frac{3 k}{14 M}} = \frac{8 M g}{3 k} \sqrt{\frac{3 k}{14 M}}$

Solution

1 At an extension of $\frac{x_{0}}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g}{10}$

2 $x_{0} = \frac{4 M g}{k}$

3 When spring achieves an extension of $\frac{x_{0}}{2}$ for the first time, the speed of the block connected to the spring is $3 g \sqrt{\frac{M}{5 k}}$

4 $a_{2} - a_{1} = a_{1} - a_{3}$

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Solution

1 At an extension of x04 of the spring, the magnitude of acceleration of the block connected to the spring is 3g10

2 x0=4Mgk

3 When spring achieves an extension of x02 for the first time, the speed of the block connected to the spring is 3gM5k

4 a2-a1=a1-a3

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1 At an extension of $\frac{x_{0}}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g}{10}$

2 $x_{0} = \frac{4 M g}{k}$

3 When spring achieves an extension of $\frac{x_{0}}{2}$ for the first time, the speed of the block connected to the spring is $3 g \sqrt{\frac{M}{5 k}}$

4 $a_{2} - a_{1} = a_{1} - a_{3}$