Question

4. A spring with a mass of 400.0 g is set into simple harmonic motion. The graph of the force of the spring vs. displacement is shown. The data of the position of the spring vs. time while it is oscillating are also shown.
Graph : https://us-static.z-dn.net/files/d48/07e4a9edd0f61852095576cb0765d951.jpg

Data : t (s) y (m)
0.0 0.00
0.2 0.09
0.4 –0.07
0.6 –0.04
0.8 0.10
1.0 –0.03
1.2 –0.08
1.4 0.09
1.6 0.01
1.8 –0.10
2.0 0.06
2.2 0.05
2.4 –0.10
2.6 0.02
2.8 0.09
3.0 –0.08
3.2 –0.03
3.4 0.10
3.8 –0.07
4.0 0.09

Do the following:
a. Determine the spring constant from the graph.
b. Make a position vs. time graph for the spring’s motion.
c. Determine the amplitude and period of the spring’s motion from your graph.
d. Calculate the period of the spring’s motion using the spring’s period equation.
e. Determine the percent error by comparing the period from the graph (experimental value) to the calculated value (accepted value).

Answer 1

4 percent i think ok because maybe i think is e

Answer 2

Answer:

a) k = 40 N/m

b) See attachment

c) Amp y = 0.1 m , T_a = 0.67 s

d) T_theo = 0.62831

e) RE = 6.63 %

Explanation:

Given:

- mass of the spring m = 0.4 kg

- Displacement - time values

- Force-displacement graph

Find:

a. Determine the spring constant from the graph.

b. Make a position vs. time graph for the spring’s motion.

c. Determine the amplitude and period of the spring’s motion from your graph.

d. Calculate the period of the spring’s motion using the spring’s period equation.

e. Determine the percent error by comparing the period from the graph (experimental value) to the calculated value (accepted value).

Solution:

a)

- Use the Force-displacement graph and pick two points on the line A and B:

                       A = ( 0.006 , 1.78 ) & B = ( 0.083, 4.75 )

- We know from Hook's Law that:

                       y = k*x

                       k = y/x       ------  gradient of the plot will give us the constant k

- So,

                       k = (y_2 - y_1) / (x_2 - x_1)=(4.75-1.78)/(0.083-0.006)=40 N/m

b) See attachment

c) Using the plot of b:

        The maximum amplitude of displacement y = 0.1 m

         The time period of the displacement T_a = 2 s / 3 cycles = 0.67s

d)

- The formula for computing time period is:

                          T_theo = 2*pi*sqrt(m/k)

                          T_theo = 2*pi*sqrt(0.4/40) = 0.62831 s

e)

- Comparing values by computing the relative error:

                          RE = ( T_a - T_theo) / T_theo * 100

                          RE = (0.67 - 0.62831) / 0.62831 * 100 = 6.63 %

- The value obtained is within the 10% tolerance of experimental Error. Hence, we can conclude that the experiment was conducted reasonably.