Question 289: Laws of Motion

Question

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 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]
![](https://imagex.cdn.examgoal.net/1kszc3ll3/d056b43c-f8cc-4952-bed5-bf2b84e0ebf2/a34ed170-09f2-11ec-837d-2903eea50152/file-1kszc3ll4.jpeg)

Accepted Answer

In the frame of pulley B,
the hanging masses have accelerations :
$M 	o ({a_2} - {a_1})$
$2M 	o ({a_3} - {a_1})$ : downward.
$ 	herefore $ $({a_2} - {a_1})$ = $( - {a_3} - {a_1})$ [constant]
Assuming that the extension of the spring is x
We consider the FBD of A :
$2M.{{{d^2}x} \over {d{t^2}}} = 2T - kx$
where, ${a_1} \equiv {{{d^2}x} \over {d{t^2}}}$ ....(i)
and the FBD of the rest of the system in the frame of pulley B :
Upward acceleration of block M w.r.t. the pulley B = Downward acceleration of block 2M w.r.t. the pulley.
${{T - M(g - {a_1})} \over M} = {{2M(g - {a_1}) - T} \over {2M}}$
$ \Rightarrow T = {{4M} \over 3}(g - {a_1})$ .....(ii)
Substituting in Eq. (i), we get
$2M.{a_1} = {{8M} \over 3}(g - {a_1}) - kx$
or ${{14M} \over 3}{a_1} = {{8Mg} \over 3} - kx$ ....(iii)
This is equation of SHM
Maximum extension = 2 $ 	imes $ amplitude
i.e., ${x_0} = 2 	imes {{8Mg} \over {3k}}$
Amplitude = ${{{x_0}} \over 2} = {{3Mg} \over {3k}}$ and $\omega  = \sqrt {{{3k} \over {14M}}} $
At ${{{x_0}} \over 4}$, acceleration is easily found from Eq. (iii),
${{14M} \over 3}{a_1} = {{8Mg} \over 3} - {{4Mg} \over 3}$
${a_1} = {{2g} \over 7}$
At ${{{x_0}} \over 2}$, speed of the block
$(2M) = \omega  	imes \,amplitude$
= $\sqrt {{{3k} \over {14M}}}  	imes {{8Mg} \over {3k}}$
$ 	herefore $ option (a) is correct.
![](['https://imagex.cdn.examgoal.net/1kszcz04j/202f1a7f-72d4-457a-a368-5e41b1f99d5f/0c9fb240-09f6-11ec-837d-2903eea50152/file-1kszcz04k.jpeg', 'https://imagex.cdn.examgoal.net/1kszd4ifv/1f14fe91-1d9a-4e46-9fcb-15428359e1a5/a5d112b0-09f6-11ec-ac1f-e362cac4ba2a/file-1kszd4ifw.jpeg', 'https://imagex.cdn.examgoal.net/1kszd7rg4/c943f044-1df6-41c6-9d48-32c1b41150da/0033a740-09f7-11ec-ac1f-e362cac4ba2a/file-1kszd7rg5.jpeg'])

Answer

${a_2} - {a_1} = {a_1} - {a_3}$

Answer

At an extension of ${{{x_0}} \over 4}$ of the spring, the magnitude of acceleration of the block connected to the spring is ${{3g} \over {10}}$.

Answer

${x_0} = {{4Mg} \over k}$

Answer

When spring achieves an extension of ${{{x_0}} \over 2}$ for the first time, the speed of the block connected to the spring is $3g\sqrt {{M \over {5k}}} $

Laws of Motion

Choose the correct answer: