Is the first graph provided in question 5  velocity vs time   or v vs position   or V vs distance? (The graph indicates that it is V vs X (the description indicates it is v vs time)   How could the position keep increasing beyond the amplitude?

in part C of the same problem the solution states that the amplitude from the graph is 0.35 m?  how is that obtained from the given graph?  If it is V vs time that information is not obtainable from the graph and if it is V vs distance travelled then the amplitude 1/4 wavelength ??????

Jacob, thanks for the input - I removed the first graph as it's not needed and as you pointed out, incorrect. I did add that the student pulled the spring out .35 m, so that would be our delta x for part c. John

John, thank you. but now information about period or velocity is necessary to get to a value for part C or delete the plugged in values in the answer for part C

Jacob, if you look at the graph at time 0, it shows a velocity of 0.16 m/s, which is the maximum velocity, before friction takes hold and slows it down. So assume that's the max velocity. This also means we have to assume that the students didn't start the clock until the pendulum reached its maximum velocity where there was no potential energy, and it was all kinetic. If they started the clock when they released the pendulum, the initial velocity would be zero. Since that assumption is not explicitly stated, I'll add it to the question. But I'd like your opinion on this before I make the change.  Thanks!  John

John,  my preference would be to add a first graph of velocity Vs position for parts a-c.

Jacob - I agree - I admit I just band aided the problem - I will work on your idea.  Thanks, John

I do not agree with the answer provided for part a of question 5.  Or may be I do not understand the question.  I think both students are correct exactly half of the time.  Whenever the block is headed toward equilibrium the spring force is in the same direction as evidenced in the fact that the velocity is increasing on the way toward equilibrium.  Student A's statement is correct for that time frame.  Student B's statement is correct when the object is headed away from the equilibrium point.

Jacob, exactly right and thank you. I added Student C with the right answer. Clearly students A and B confused the direction of the force with the direction of the motion - rookie physics mistake.  John

The answer to Question #3 part c is VERY wrong.  The greatest tension is achieved AT the equilibrium point.  The reason being that tension at the equilibrium point is mg + mv^2/r and v is a maximum at the equilibrium point.  At the amplitude points, the acceleration is purely tangential (velocity is zero, so centripetal acceleration is 0).  Rotating the axis like with inclines, the tension force then becomes equal to mg (cos theta) where theta is the angle with the vertical.