# Spring pulley system

Hey can someone please help me with this question, the answer given is B and C, but my doubt is that the acceleration of both the masses should not be the same and $K_1x_1 = K_2x_2$ ...[equation 1] where $K_1$ and $K_2$ are the spring constants and $x_1$ and $x_2$ are the elongations of the respective springs.

If we differentiate the equation twice(with respect to time) then $K_1a_1 = K_2a_2$, and thus unless $K_1=K_2$ the accelerations shouldn't be the same (in the solution given they have assumed the accelerations to be the same).

Also the system should never be able to reach equilibrium as in equilibrium $K_1x_1 =m_1g$, $K_2x_2=m_2g$ ($m_1$ and $m_2$ are masses of the respective objects) which implies $m_1g=m_2g$ (from equation 1),so how can the elongation be a constant and independent of $x_2$?

As Farcher points out, the system is subject to a constant unbalanced force, so the resulting motion is constant acceleration. It is never in equilibrium : it is neither static nor moving with constant speed.

Your eqn 1 is correct : $K_1x_1=K_2x_2$. However, the tension in the springs is constant, so $x_1$ and $x_2$ are constants, and when you differentiate they vanish. $x_1$ and $x_2$ are not measures of the positions of the two masses, so differentiating them does not give you the linear acceleration of each mass.

Perhaps your difficulty is caused by the ideal conditions required for the two masses to move with constant acceleration. Unless the masses are released very carefully, and the pulley rotates very smoothly, any small jerk in the motion will be amplified by the springs : tension and extension will vary with time. The resulting motion of each mass will not be constant acceleration, even if the motion of the $CM$ is. The 'solution' of constant acceleration is an unstable one. I find it intuitive that the masses will oscillate as the heavier one descends.

• But initially the individual masses will move with non constant accelerations as the springs will keep elongating,and eventually they will move with a constant acceleration once the elongation becomes a constant, right? – user329387 Apr 24 '16 at 4:32
• I think it must be assumed that the system is released when the tension in the springs is the constant value in Farcher's calculation. It will then undergo constant acceleration. You are correct to suppose that, if released when x1=x2=0, there will be an initial period of non-constant acceleration (oscillations). Depending on the particular values of m and K, these oscillations may die out quickly or be amplified. The problem addresses an ideal situation rather than a practical one. – sammy gerbil Apr 24 '16 at 15:58

Draw an free body diagram for each of the mass, apply N2L, eliminate the acceleration $a$ to find the tension $T$.

• Ya but why will the accelerations of both the masses be the same? And how can the system ever reach equilibrium? – user329387 Apr 21 '16 at 15:19
• The system never reaches equilibrium. So the tension in the string is the force applied to each end of the spring. $T =kx$ – Farcher Apr 21 '16 at 15:20
• But then the elongation that we will obtain would be the elongation at which instant? As the system would never reach equilibrium, the elongation should depend on x2 right? – user329387 Apr 21 '16 at 15:23
• If the acceleration is constant the tension in the string must be constant and so the extension of the springs is constant. Follow through with the equations that I have given you and see what you get. – Farcher Apr 21 '16 at 15:35
• But the accelerations aren't constant either because as either mass moves down, the spring would elongate more and thus exert a greater force on the mass – user329387 Apr 21 '16 at 15:37