Block and Spring

Classical Mechanics Level 3


When the arrangement shown is in equilibrium, the spring on the right is stretched by an amount x 0 . x_{0}.

Coefficient of static friction between the blocks is μ \mu and the horizontal floor is frictionless. Both the blocks have equal mass m m and the horizontal floor is frictionless. Both the blocks have equal mass m m and force constants of the spring are 3 k 3k and k k as shown in the figure.

The maximum amplitude of oscillations of the blocks along with the springs that do not allow them to slide on each other is in the form α μ m g k + β x 0 , \frac{ \alpha \mu mg}{k} +\beta x_{0}, where the acceleration of free fall is g g .

What is the sum of constants α + β \alpha + \beta .


The answer is 0.

Talulah Riley by Talulah Riley , 18, India

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1 solution

What is to be submitted as the answer is not given. Let us assume that it is α + β α+β .

Initially it is not told whether the spring on the left has any stretch or not.

Equilibrium condition is : Applied force ≤ \leq limiting value of friction force . It is not told whether the system impends motion initially (the two forces are equal). Let us assume that they are equal : k x 0 = μ m g kx_0=\mu mg

Let the maximum stretch of the right spring, which determines the amplitude required, be x x . Then

3 k ( x − x 0 ) + μ m g = k x ⟹ 3k(x-x_0)+\mu mg=kx\implies

x = 3 x 0 2 − μ m g 2 k x=\dfrac {3x_0}{2}-\dfrac {\mu mg}{2k}

So α = − 1 2 , β = 1 2 , α=-\dfrac 12,β=\dfrac 12, and α + β = 0 α+β=\boxed 0 .

1 Helpful 1 Interesting 1 Brilliant 0 Confused

Nice, that's how I did it too.

Krishna Karthik - 7 months, 1 week ago

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