Defining Simple Harmonic Motion (SHM) AP Physics 1 FRQ – Exam Style Questions etc.
Defining Simple Harmonic Motion (SHM) AP Physics 1 FRQ
Unit 7: Oscillations
Weightage : 10-15%
Exam Style Practice Questions, Defining Simple Harmonic Motion (SHM) AP Physics 1 FRQ
Question
A group of students have been asked to determine the spring constant k of the following setup. They have a known mass and a very smooth tabletop. The only additional equipment they have is metersticks and rulers.
(a) Describe an experimental procedure to determine the spring constant k . You may wish to label the diagram above further to help in your description. Indicate what measurements you would take and how you would take them. Include enough detail so that another student could carry out your procedure.
(b) What are the expected results of the experiment? Determine your independent and dependent variables. Sketch out the expected graph. Derive the relationship you would expect to find.
(c) What common sources of error in your procedure might happen during this investigation? What steps could you take to minimize these errors?
(d) A group of students collected the following data; x is for the spring and D is along the ground. The students found a discontinuity in the behavior of the graph. Graph the data and determine a value for k . Offer a possible explanation for the discontinuity.
Answer/Explanation
Ans:
(a) For various measured compressions of the spring (x), measure horizontal range for the mass (R). Range should be measured along the floor from beneath the edge of the table to where the mass first hits the ground. Multiple trials for each compression x should be taken so that the average range of values can be determined.
(b) The independent variable is the one the experimenter controls and manipulates directly. In this case, the independent variable is the compression x . The dependent variable is the one measured as a result of changes in the independent variable. In this case, the dependent variable is the range. Independent variables are graphed on the horizontal axis. Theoretical prediction:
\(\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}\)
\(v=(k/m)^{1/2}x\)
This velocity is the horizontal projectile’s velocity. The time in flight is found from the height of the table:
\(H=\frac{1}{2}gt^{2}\)
\(1.6m=4.9t^{2}\)
\(t = 0.57s\)
\(R=vt=(k/m)^{1/2}xt=.57(k/m)^{1/2}x\)
Using the fixed values for m and t: \(R=0.806k^{1/2}x\)
(c) The major source of error is friction of the tabletop between the end of the spring and the edge of the table. Ensuring that this distance is small and that the surfaces involved are smooth will minimize this error.
(d) 
The discontinuity is probably caused by exceeding the limit of elasticity for this spring. Hooke’s law assumes that the material is perfectly elastic and resumes its shape after being stretched or compressed:
Slope = 20 = 0.806 \(k^{1/2}\) (from part (b))
Solving for k: \(k\) = 616 N/m
Question: (12 points, suggested time 25 minutes)
A block is initially at position x = 0 and in contact with an uncompressed spring of negligible mass. The block is pushed back along a frictionless surface from position x = 0 to x = -D , as shown above, compressing the spring by an amount Δx = D . The block is then released. At x = 0 the block enters a rough part of the track and eventually comes to rest at position x = 3D . The coefficient of kinetic friction between the block and the rough track is μ .
(a) On the axes below, sketch and label graphs of the following two quantities as a function of the position of the block between x = -D and x = 3D . You do not need to calculate values for the vertical axis, but the same vertical scale should be used for both quantities.
i. The kinetic energy K of the block
ii. The potential energy U of the block-spring system
The spring is now compressed twice as much, to Δx = 2D . A student is asked to predict whether the final position of the block will be twice as far at x = 6D . The student reasons that since the spring will be compressed twice as much as before, the block will have more energy when it leaves the spring, so it will slide farther along the track before stopping at position x = 6D .
(b)
i. Which aspects of the student’s reasoning, if any, are correct? Explain how you arrived at your answer.
ii. Which aspects of the student’s reasoning, if any, are incorrect? Explain how you arrived at your answer.
(c) Use quantitative reasoning, including equations as needed, to develop an expression for the new final position of the block. Express your answer in terms of D.
(d) Explain how any correct aspects of the student’s reasoning identified in part (b) are expressed by your mathematical relationships in part (c). Explain how your relationships in part (c) correct any incorrect aspects of the student’s reasoning identified in part (b). Refer to the relationships you wrote in part (c), not just the final answer you obtained by manipulating those relationships.
Answer/Explanation
Ans:
(a) 
(b)
i. The student is correct to say that any eassing nor spring will result more a longer kinetic energy value for the block when it leave the spring. Springs store more energy when compressed further and that greater amount of energy will be transferred to the block.
ii. The student B me or not to suppose that the block will go twice as far. The amount of energy shared in a spring B proportioned to the amount of is compressed squared, Because it is compressed there as far, it will have firmes the potential energy which will be transferred to the block, which will also have foretimes as much energy and will go four times as far.
(c) d= final position
EK = EU EU= Ff . d = mgμ. d \(d_{1}=\frac{k}{mg\mu }.D^{2} = 3D\)
\(E_{k}=\frac{1}{2}kx^{2}\) \(\sum Fy = F_{N}-mg=0\)
\(F_{f}=mg\mu \) \(d_{2}=\frac{k(2D)^{2}}{mg\mu }=4\left ( \frac{KD^{2}}{mg\mu } \right )\)
= 4 (3D)
\(\frac{1}{2}kx^{2}= mg\mu .d\) \(D = \frac{kx^{2}}{mg\mu }\) = 12 D
Four times the energy will be stored in the spring, four times the energy will go into the block, and fraction will take 4 times the distance to disperse the energy.
(d) More potential energy will be stored on the spring when it is compressed further as the student stated and as illustrated by the relationship \(E_{u}=\frac{1}{2}k\Delta x^{2}.\) This relationship also shows that when the spring is compressed twice as far (Δx B doubled) four times no energy is stored in the spring. The force of fraction is the same in both instances, so the same amount of energy is dissipated by friction per unit of length translated by E=Ff.d) so 4 times the energy will take 4 times the distance to be dissipated, not twice the distance as the student who assumed that potential energy is proportional to ΔX supposed.