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Text assignments:

General College Physics:  Chapter 11, Sections 1-4.  Do problems 7, 14, 24, 32, 33

University Physics: As on Assignment page for Chapter 9.

What is the KE of a simple harmonic oscillator at x = 15 cm, given that the restoring force constant is k = 120 N / m?

PE = .5 k x^2 = .5 ( 120 N / m) ( .15 m)^2 = 1.35 J.

What is the KE of your pendulum halfway to its equilibrium position, and what should be the horizontal range of the dime if it is released from rest and stopped precisely at this position?

What is PE at maximum displacement?

For a pendulum k = m g / L

PE = .5 k x^2.

What x = A we have PE = .5 k A^2.

So we use your given k and A to find PEmax = .5 k A^2.

Then we find PEhalfway = .5 k ( A / 2)^2.

Assuming no significant dissipative forces, KE change is the negative of PE change so KE change is PEmax - PEhalfway.

How do we find the velocity at the halfway point?

Set .5 m v^2 equal to KE and solve for v.  You know m.

When we stop the pendulum, the dime will come off with this initial velocity, in a direction perpendicular to the string.

A little trigonometry tells us that the direction is arcTan(x / sqrt(L^2-x^2)), where x is the displacement from equilibrium and L the length of the pendulum.

The direction with respect to the standard coordinate plane is

Figure out this angle for your displacement and figure out your initial vertical and horizontal velocities.

Predict the range of the dime from this position.

To predict the range we solve the equations of motion for the vertical motion of the dime obtaining time of fall.

We then find the horizontal displacement from the constant horizontal velocity and time of fall.

length vMax pred range obs range   max PE halfway PE halfway KE halfway v est range theta v0y v0x vfy dt pred range
9 26 14.6 18   0.034 0.008481 0.0254423 0.22558 0.1269 2.2053 -0.009 0.225 -5.51181 0.56154 0.12658
12 34.7 19.5 25   0.06 0.015077 0.0452308 0.30077 0.1692 2.941 -0.015 0.3 -5.51183 0.56086 0.16847
15 43.4 24.4 31   0.094 0.023558 0.0706731 0.37596 0.2115 3.6772 -0.024 0.375 -5.51186 0.55998 0.2101
18 52.1 29.3 19   0.136 0.033923 0.1017692 0.45115 0.2537 4.414 -0.035 0.45 -5.51191 0.5589 0.2514
21 60.8 34.2     0.185 0.046173 0.1385192 0.52634 0.296 5.1515 -0.047 0.524 -5.51201 0.55763 0.29232
24 69.5 39.1 45   0.241 0.060308 0.1809231 0.60154 0.3383 5.8898 -0.062 0.598 -5.51215 0.55617 0.33279
27 78.1 43.9 44   0.305 0.076327 0.2289808 0.67673 0.3806 6.6292 -0.078 0.672 -5.51236 0.55452 0.37275
30 86.8 48.8 59   0.377 0.094231 0.2826923 0.75192 0.4229 7.3696 -0.096 0.746 -5.51265 0.55268 0.41214
33 95.5 53.7 60   0.456 0.114019 0.3420577 0.82711 0.4652 8.1113 -0.117 0.819 -5.51304 0.55065 0.4509

Add or subtract 1 cm on total pullback and predict range, and predict the difference the 1 cm change makes in the range.

University Physics students:  Also correct for the fact that the height at which the dime slides off is a little different than the 155 cm we have at equilibrium.