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QJEEAdv19Ph030e: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ph030e Solution: A block of mass 2M is attached to a massless spring with…

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 a1, a2 and a3 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 x0. Which of the following option(s) is/are correct? [g is the acceleration due to gravity. Neglect friction] JEE Advanced 2019 Paper 2 Offline Physics - Laws of Motion Question 12 English

Multiple Choice Options

  • A a2−a1=a1−a3{a_2} - {a_1} = {a_1} - {a_3}
  • B At an extension of x04{{{x_0}} \over 4} of the spring, the magnitude of acceleration of the block connected to the spring is 3g10{{3g} \over {10}} .
  • C x0=4Mgk{x_0} = {{4Mg} \over k}
  • D When spring achieves an extension of x02{{{x_0}} \over 2} for the first time, the speed of the block connected to the spring is 3gM5k3g\sqrt {{M \over {5k}}}

Step-by-Step Verified Solution

JEE Advanced 2019 Paper 2 Offline Physics - Laws of Motion Question 12 English Explanation 1

In the frame of pulley B,

the hanging masses have accelerations :

M→(a2−a1)M \to ({a_2} - {a_1})

2M→(a3−a1)2M \to ({a_3} - {a_1}) : downward.

∴\therefore (a2−a1)({a_2} - {a_1}) = (−a3−a1)( - {a_3} - {a_1}) [constant]

Assuming that the extension of the spring is x

We consider the FBD of A :

2M.d2xdt2=2T−kx2M.{{{d^2}x} \over {d{t^2}}} = 2T - kx

JEE Advanced 2019 Paper 2 Offline Physics - Laws of Motion Question 12 English Explanation 2

where, a1≡d2xdt2{a_1} \equiv {{{d^2}x} \over {d{t^2}}} ....(i)

and the FBD of the rest of the system in the frame of pulley B :

JEE Advanced 2019 Paper 2 Offline Physics - Laws of Motion Question 12 English Explanation 3

Upward acceleration of block M w.r.t. the pulley B = Downward acceleration of block 2M w.r.t. the pulley.

T−M(g−a1)M=2M(g−a1)−T2M{{T - M(g - {a_1})} \over M} = {{2M(g - {a_1}) - T} \over {2M}}

⇒T=4M3(g−a1)\Rightarrow T = {{4M} \over 3}(g - {a_1}) .....(ii)

Substituting in Eq. (i), we get

2M.a1=8M3(g−a1)−kx2M.{a_1} = {{8M} \over 3}(g - {a_1}) - kx

or 14M3a1=8Mg3−kx{{14M} \over 3}{a_1} = {{8Mg} \over 3} - kx ....(iii)

This is equation of SHM

Maximum extension = 2 ×\times amplitude

i.e., x0=2×8Mg3k{x_0} = 2 \times {{8Mg} \over {3k}}

Amplitude = x02=3Mg3k{{{x_0}} \over 2} = {{3Mg} \over {3k}} and ω=3k14M\omega = \sqrt {{{3k} \over {14M}}}

At x04{{{x_0}} \over 4} , acceleration is easily found from Eq. (iii),

14M3a1=8Mg3−4Mg3{{14M} \over 3}{a_1} = {{8Mg} \over 3} - {{4Mg} \over 3}

a1=2g7{a_1} = {{2g} \over 7}

At x02{{{x_0}} \over 2} , speed of the block

(2M)=ω× amplitude(2M) = \omega \times \,amplitude

= 3k14M×8Mg3k\sqrt {{{3k} \over {14M}}} \times {{8Mg} \over {3k}}

∴\therefore option (a) is correct.

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