Spring 26-09-2020

Classical Mechanics Level pending

A block of mass m m placed on a frictionless horizontal floor is connected with two identical springs each of force constant k k .
One end of the left spring is connected to a fixed support and one end of the right spring is free. Initially the block is at rest, the springs are collinear and relaxed. If someone begins to pull the free end of the right spring with a constant velocity v v away from the wall, how far will the block move before it acquires a speed of v v .

Answer comes in the form of α v m k \alpha v \sqrt{\frac{m}{k}}
Type α \alpha


The answer is 1.110.

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1 solution

Karan Chatrath
Sep 26, 2020

Nice problem. Let the left end of the left spring be fixed at the origin. The X coordinate of the mass is x 1 x_1 and the X coordinate of the rightmost point of the right spring is x 2 x_2 . Say both springs have a natural length of L o L_o . We know that:

x ˙ 2 = v \dot{x}_2 = v x 2 ( 0 ) = 2 L o x_2(0) = 2L_o

x 2 = v t + 2 L o \implies x_2 = vt+2L_o

Now, the kinetic energy of the system is:

T = m x ˙ 1 2 2 T = \frac{m\dot{x}_1^2}{2} V = k 2 ( ( x 2 x 1 L o ) 2 + ( x 1 L o ) 2 ) V = \frac{k}{2}\left((x_2 - x_1 -L_o)^2 + (x_1 - L_o)^2\right)

Plugging in the expression for x 2 x_2 and substituting x = x 1 L o x = x_1 - L_o gives:

T = m x ˙ 2 2 T = \frac{m\dot{x}^2}{2} V = k 2 ( ( v t x ) 2 + x 2 ) V = \frac{k}{2}\left((vt-x)^2 + x^2\right)

x ( 0 ) = x ˙ ( 0 ) = 0 x(0) = \dot{x}(0) = 0

Now, Lagrangian Mechanics can conveniently handle the rest. I am leaving out this step and presenting the equation of motion, which can also be easily derived using Newton's law (The reader may try this by him/her self):

m x ¨ + 2 k x = k v t m\ddot{x} + 2kx = kvt x ( 0 ) = x ˙ ( 0 ) = 0 x(0) = \dot{x}(0) = 0

The solution of this equation is (again, I have left out the steps):

x = v 2 ( t sin ( ω t ) ω ) x = \frac{v}{2}\left(t - \frac{\sin(\omega t)}{\omega}\right) ω 2 = 2 k m \omega^2 = \frac{2k}{m}

Now,

x ˙ = v 2 ( 1 cos ( ω t ) ) \dot{x} = \frac{v}{2}(1 - \cos(\omega t))

x ˙ = v \dot{x} = v only when cos ( ω t ) = 1 \cos(\omega t) = -1 . Therefore:

t = π ω t = \frac{\pi}{\omega}

Plugging this into the solution for x x gives:

x = v π 2 ω x = \frac{v \pi}{2 \omega} x = ( π 2 2 ) v m k x = \left(\frac{\pi}{2\sqrt{2}}\right)v\sqrt{\frac{m}{k}} α = π 2 2 \therefore \alpha = \frac{\pi}{2\sqrt{2}}

@Karan Chatrath nice solution upvoted.

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath Thanks for the solution

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath Now I understand the solution. Yeah. Cheers!!!

Talulah Riley - 8 months, 2 weeks ago

@Karan Chatrath You solved it so fastly, very nice.

Talulah Riley - 8 months, 2 weeks ago

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