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]

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.

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: