a.) We know the spring constant and the oscillation frequency. So we can solve for the mass
b.) The force on the mass is equal to -ky, where y is the displacement. If the displacement is one-half of its maximum, then it's equal to 2.5 cm = 0.025 m (since the maximum displacement is the same as the amplitude). This means the force there equals -(250 N/m)(0.025 m) = -6.2 N. The minus sign means that the force pulls the mass towards equilibrium if the spring is stretched.
c.) Looking at the position versus time and the velocity versus time graphs, you should be able to see that the slope of lines tangent to the velocity versus time graph are steepest when the mass is at it's maximum displacement.
d.) All you can do to change the frequency of oscillation is to replace the spring or replace the mass. Using the formula from a.), you need a spring with force constant 4 times smaller or a mass that is 4 times bigger.
a.) The mass returns to equilibrium in each half cycle. This means that the first time it returns to equilibrium is in T/2, where T is the period and is T = 1/f. Using the formula from S14 a.), we calculate the frequency to be 2.25 Hz, meaning the period is T = 1/f = 0.44 s. The mass first returns to equilibrium in a time T/2 = 0.22 s.b.) It takes a quarter cycle more (0.11 s) for the mass to reach its maximum negative displacement. It gets there for the first time in 3T/4 = 0.33 s. The time to get there the second time is obtained by adding a full period; so, the total time is 0.77 s.
c.) To get the position after 2.0 seconds, we need to evaluate the sin function using a period, T = 0.44 s.
d.) The maximum velocity (always occuring at the equilibrium position) is given in terms of the amplitude and period as vmax = 2pi ymax / T = (6.28)(0.05 m) / (0.44 s) = 0.71 m/s. When the mass first returns to the equilibrium position, it is going in the other direction or v = -vmax = -0.71 m/s
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Last updated: 16 Sep 1999
Comments: bland@indiana.edu