Representing and Analyzing SHM AP Physics 1 FRQ – Exam Style Questions etc.
Representing and Analyzing SHM AP Physics 1 FRQ
Unit 7: Oscillations
Weightage : 10-15%
Exam Style Practice Questions, Representing and Analyzing SHM AP Physics 1 FRQ
Question
An experiment is designed to calculate the spring constant k of a vertical spring for a jumping toy. The toy is compressed a distance of x from its natural length of \(L_{0}\), as shown on the left in the diagram, and then released. When the toy is released, the top of the toy reaches a height of h in comparison to its previous height and the spring reaches its maximum extension. The experiment is repeated multiple times and replaced with different masses m attached to the spring. The spring itself has negligible mass.
(a) Derive an expression for the height h in terms of m, x, k, and any other constants provided in the formula sheet.
(b) To standardize the experiment, the compressed distance x is set to 0.020 m. The following table shows the data for different values of m.
(i) What quantities should be graphed so that the slope of a best-fit straight line through the data points can help us calculate the spring constant k?
(ii) Fill in the blank column in the table above with calculated values. Also include a header with units.
(c) On the axes below, plot the data and draw the best-fit straight line. Label the axes and indicate scale.
(d) Using your best-fit line, calculate the numerical value of the spring constant.
(e) Describe an experimental procedure that determines the height h in the experiment, given that the toy is only momentarily at that maximum height. You may include a labeled diagram of your setup to help in your description.
Answer/Explanation
Ans:
(a) The spring observes Conservation of Mechanical Energy
\(mgh=\frac{1}{2}kx^{2}\)
\(h=\frac{kx^{2}}{2mg}\)
(b) (i) The value of 1/m would help us in determining the spring constant when you take the slope of the line.
(ii) 
(c) 
(d) To calculate the slope, select two points on our best-fit line. Here is one
\(slope=\frac{(0.42-0.10)m}{(40-10)kg^{-1}}=\frac{0.32m}{30kg^{-1}}=1.07\times 10^{-2}m\cdot kg\)
From part (a), \(h=\frac{kx^{2}}{2mg}\rightarrow slope=\frac{kx^{2}}{2g}\)
Note: Values between 450 N/m and 550 N/m are acceptable.
\(k=\frac{2g(slope)}{x^{2}}=\frac{2(9.8m/s^{2})(1.07\times 10^{-2}m\cdot kg))}{(0.02m)^{2}}=524N/m\)
(e) One example would be to use a meter stick and video camera. Hold the meter stick next to the toy and allow it to jump up. Record the toy with a video
camera. With this you can watch the video in slow motion later to determine its height. You can also do an alternate method of finding the height by recording the time t it takes to fall from its highest point to the lowest. Then, using kinematics equations, you can determine the height.
Question
A horizontal spring with a force constant of 40 N/m is attached a 0.1 kg block.
(a) If the block is pulled to a distance of 0.5 m and released, what is the maximum speed of the block?
(b) What is the frequency of the oscillations?
(c) If the spring were flipped vertically and attached to the ground with the block placed on top, how would the natural length of the spring change?
(d) How does the frequency of the oscillations of the vertical spring-block oscillator compare with the frequency when it was placed horizontally?
Answer/Explanation
Ans:
(a) When pulled to a distance of 0.5 m, the block has maximum potential energy given by the equation \(U=\frac{1}{2}kA^{2}\). All of this energy is converted to kinetic energy when the block reaches maximum speed:
(b) The frequency of oscillations is given by the following equation:
\(f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{40N/m}{0.1kg}}=\frac{10}{\pi }Hz=3.2Hz\)
(c) When the spring is placed vertically on the ground with the block on top, there are two forces acting on the block at the new natural length: the weight of the block and the upwards restoring force. These two forces are equal to each other at the natural length:
\(F_{rest}=F_{g}\)
\(kx = mg\)
\(x=\frac{mg}{k}=\frac{(0.1kg)(10m/s^{2})}{40N/m}=0.025m\)
The natural length of the spring would decrease by 0.025 m
(d) The frequency of spring-block oscillator is given by the equation \(f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}\). Flipping the spring vertically does not affect k or m, so the frequency of oscillations will remain unchanged.