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

(C) $\mathrm{a}_{2}-\mathrm{a}_{1}=\mathrm{a}_{1}-\mathrm{a}_{3}$

(A) $\mathrm{x}_{0}=\frac{4 \mathrm{Mg}}{\mathrm{k}}$

(B) When spring achieves an extension of $\frac{\mathrm{x}_{0}}{2}$ for the first time, the speed of the block connected to the spring is $3 \mathrm{~g} \sqrt{\frac{\mathrm{M}}{5 \mathrm{k}}}$

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

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Laws of Motion

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A block of mass 2M is attached to a massless spring with spring-constant k . This block is connected to two other blocks of masses M and 2M using two massless pulleys and strings. The accelerations of the blocks are

a_{1}

a_{2}

and

a_{3}

as shown in 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]

(A) $x_{0} = \frac{4 Mg}{k}$
(B)When spring achieves an extension of $\frac{x _{0}}{2}$ for the first time, the speed of the block connected to the spring is $3 g \frac{M}{5 k}$
(C) $a_{2} - a_{1} = a_{1} - a_{3}$ ✓ Correct
(D)At an extension of $\frac{x _{0}}{4}$ of the spring, the magnitude of acceleration of the block connected to the spring is $\frac{3 g}{10}$

Answer: (C)

Solution

From constraint relations:

2 a_{1} = a_{2} + a_{3}

, which gives

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

. Option (C) is correct.

Analyzing forces on the 2M block:

\frac{8 M g}{3} - k x = 2 M a_{1}

This can be rewritten as:

a_{1} = - \frac{3 k}{14 M} [x - \frac{8 M g}{3 k}]

This is SHM about

x_{1} = \frac{8 M g}{3 k}

with

ω = \frac{3 k}{14 M}

.

Maximum extension:

x_{0} = 2 x_{1} = \frac{16 M g}{3 k}

, not

\frac{4 M g}{k}

. Option (A) is incorrect.

At

x = x_{0} /2

, the block passes through mean position with speed:

v_{0} = A ω = \frac{8 M g}{3 k} \cdot \frac{3 k}{14 M} = \frac{62 M g ^{2}}{21 k}

This doesn't match option (B), which is incorrect.

At

x = x_{0} /4

a_{1} = \frac{8 M g - 3 k ( x _{0} /4 )}{14 M} = \frac{2 g}{7}

This is not

\frac{3 g}{10}

. Option (D) is incorrect.

JEE Advanced · 2019 · 27 May 2019 · Paper 2 · One or more correct options

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JEE Advanced · 2019 · 27 May 2019 · Paper 2 · Laws of Motion

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