The answer is C, but why? At first I thought it was B because the springs would shake the block causing it to produce a wave and as more time has elapsed there would have been more time for friction to act on the block making the block slowly returning to equilibrium. My second possible reasoning that is assuming the springs both have the same spring constant the block would remain at rest because the force each spring exerts is equal and opposite to the other spring resulting in nothing happening.
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2 Answers
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The picture you see is called a wave profile, a photograph (snapshot) of the wave at one instant of time.
The wave is travelling from left to right which means that the part of the wave on the right was formed by the motion of the block at an earlier time than the time at which the photograph was taken.
At that time in the past the amplitude of the block's oscillation was larger than at the time the photograph was taken and hence the amplitude of the wave is larger at the right hand side.
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$\begingroup$ Thanks for explaining but I just want to clarify, are you proposing that the waves produced by the object was initially B then slowly transformed into C as friction had more time to act on the system? $\endgroup$– coderhkCommented Apr 28, 2017 at 5:31
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$\begingroup$ When the mass was first released it was displaced a certain amount & so the string was displaced by that distance. There were no other waves on the string. At time progressed the more waves of smaller amplitude where produced and those which were produced earlier had moved down the sting. $\endgroup$– FarcherCommented Apr 28, 2017 at 5:40
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$\begingroup$ If I was to label the wave profile as if it were a graph representing the motion of the block as a function of time the vertical axis would be displacement of the block. The horizontal axis would be time with, say, the left hand being labelled zero meaning at the time the photograph was taken and the times becoming more negative (i.e. position of block at times before the photograph was taken) as one progressed to the right. $\endgroup$– FarcherCommented Apr 28, 2017 at 5:44
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I believe you have a small logical mistake: The blocks having the same constant (which is actually not necessary to consider here anyway) wouldn't lead to the block staying at rest. The springs follow the differential force equation which allows them to oscillate. Now, from rest they wouldn't start oscillating but here a small variation is given (at time 0 it's not at rest). But again using the force equation we find that the oscillation is damped. Now the string that is attached basically just shows you a few of the last oscillations (moving from left to right) and as their amplitudes get smaller C is correct as you already knew.
The force equation i mentioned before (for one spring, without friction) would be: $m\ddot x = k x$ where the right side is hook's law.
