When two blocks match velocities

First hint. Imagine tying the two blocks together. What force would the lower block need from its neighbour, and where in the oscillation would it need the most?

Second hint. During a slip, sketch both velocities on the same axes. At their first intersection, ask what force would keep their slopes equal.

The threshold for slipping

Take right as positive. Write \(x\) for the upper block’s displacement from the position of zero spring extension, and \(X\) for the lower block’s displacement from its release position. If the blocks move together, the spring accelerates their combined mass. The only horizontal force on the lower block is friction, so the friction it needs is

The upper block presses on the lower one with a normal force \(mg\). Write \(f_{0} = \mu mg\) for the greatest available friction. The demand peaks at the turning points, where \(|x| = A\). Equating that demand to \(f_{0}\) gives

At or below this amplitude, static friction can sustain the common oscillation of angular frequency \(\Omega = \sqrt{k/(m + M)}\). From the right turning point to the centre, friction supplies energy to the lower block at exactly the rate it takes energy from the upper block. Their common contact velocity makes the net frictional power zero.

The first slip

Above the threshold, the upper block starts sliding left across the lower one. Friction now acts right on \(m\) and left on \(M\). Put \(b = f_{0}/k\), \(a_{0} = f_{0}/M\) and \(\omega = \sqrt{k/m}\). The spring and kinetic friction together exert the force \(- k(x - b)\) on the upper block. It oscillates about \(b\) as long as this direction of slip persists. The lower block accelerates uniformly. With \(\theta = \omega t\) their motions are

Both velocity curves start at zero. The spring makes the upper block’s velocity more negative at first, but its acceleration will change continuously. The lower block’s velocity keeps falling along a straight line while the two surfaces slide in the same relative direction. A reversal of the upper block’s acceleration alone would not end the slip.

Following these equations beyond the first return to \(w = 0\) would keep applying the old friction force after its physical justification had expired.

What happens where the velocities meet

The first slip ends when \(w\) returns to zero. In the first-slip velocity equations, a sine curve meets a straight line. Their intersection at the first positive angle \(\theta_{e}\) satisfies

Since \(A > A_{c}\), we have \(0 < \rho < 1\). On \(0 < \theta < \pi\), the ratio \(\sin\theta/\theta\) falls from 1 to 0. There is exactly one crossing there, before the upper block reaches its left turning point.

The plot uses equal masses and \(A = 10A_{c}\). The grey dashed line continues the old equation beyond its range of validity. What the blocks actually do there depends on the force they need.

At the crossing, \(t_{e} = \theta_{e}/\omega\). Both blocks move left with velocity \(v_{e} = - a_{0}t_{e}\), and \(x_{e} = b + (A - b)\cos\theta_{e}\). Test the friction for common acceleration again. It is available when \(|x_{e}| \leq A_{c}\). The sign of the common velocity settles what happens at the boundary.

Inside the interval \(|x_{e}| < A_{c}\), the blocks stick for at least a while. The resulting common oscillation would have amplitude

If \(B > A_{c}\), common motion lasts only until the blocks reach the boundary on their way out. A pair at the right boundary moving left would enter the allowed interval. At \(x_{e} = - A_{c}\) the blocks are already moving outwards to the left. No interval of sticking can begin. For \(x_{e} < - A_{c}\) the spring pushes the upper block right too strongly for friction to hold the pair together. In both cases the upper block starts sliding right relative to the lower one, even while both still travel left.

A second threshold to find

Try equal masses. Is there a particular initial pull at which the first catch changes from temporary sticking to immediate reversal? Eliminate \(A - b\) using the crossing equation. Since \(A_{c} = 2b\), the test takes the form

As the pull grows, \(\theta_{e}\) increases towards \(\pi\) and this expression decreases. The boundary \(x_{e} = - A_{c}\) gives \(\theta_{e}\cot\theta_{e} = - 3\). Its root is about \(2.456\) radians, corresponding to \(A \simeq 2.438A_{c}\). Just below that second threshold the blocks stick for a finite time. At the threshold itself they immediately reverse their relative sliding.

What a large pull changes

For \(A \gg A_{c}\), the first catch occurs close to a half period of the upper block. Setting \(\theta_{e} = \pi - \varepsilon\) in the crossing equation gives \(\varepsilon \simeq \pi\rho\). The lower block’s speed peaks locally at the reversal. Its exact value there is \(a_{0}\theta_{e}/\omega\). Over the first oscillation its velocity is approximately triangular, with acceleration \(- a_{0}\) and then \(+ a_{0}\) over successive half periods. Thus the greatest speed to leading order is

What sets the energy loss? The frictional forces have opposite signs, so their powers sum to \(- f_{0}|\dot{x} - \dot{X}|\) during sliding. Heat therefore records the distance one surface travels relative to the other. Over a large oscillation the lower block travels little compared with \(A\). The relative sliding distance is approximately \(4A\). Let \(\Delta A\) be the decrease between successive right turning points. To leading order,

Doubling a sufficiently large initial pull doubles the heat released in the first cycle. It also doubles the energy cost per unit decrease of amplitude. Those two factors cancel. The same estimate loses accuracy as the motion approaches the sticking threshold.

One numerical check

Take \(m = M = 1.00\, kg\), \(k = 20.0\, N\, m^{- 1}\), \(\mu = 0.200\), \(g = 9.81\, m\, s^{- 2}\) and \(A = 10A_{c}\). The exact first catch gives \(t_{e} = 0.6672\, s\) and lower block speed \(1.309\, m\, s^{- 1}\), against the half-period estimate of \(1.378\, m\, s^{- 1}\).

At this catch, \(x_{e} = - 1.743\, m\) and \(X_{e} = - 0.4367\, m\). The upper block has slid left across the lower one by \(A - x_{e} + X_{e}\). Multiplying by \(f_{0}\) gives \(6.412\, J\) of heat, using unrounded values. Independently the system’s mechanical energy falls from \(38.494\, J\) to \(32.083\, J\), the same loss within rounding.

How different are the arrows? Immediately after this catch the lower block accelerates right at \(f_{0}/M = 1.96\, m\, s^{- 2}\). The upper block also accelerates right, at \(\left( - kx_{e} - f_{0} \right)/m = 32.9\, m\, s^{- 2}\). Both blocks are still travelling left.

Now change just the initial pull to \(1.10A_{c}\). The same crossing equation gives \(x_{e} = 0.811A_{c}\) and \(B = 1.088A_{c}\). This time the arrows coincide for a while. As the pair moves left through the centre and reaches \(- A_{c}\), the common acceleration again demands more friction than is available just beyond that point. The sketch that answered the first catch cannot settle the rest of the oscillation.