10.2. Mass on a Spring http://www.ck12.org
equilibrium is the same time it takes the object to travel one-fourth of a revolution about a circle. Click on the link
to see this idea animated.
http://en.wikipedia.org/wiki/Simple_harmonic_motion
T
4
→
1. 62 s
4
= 0. 405 sii. Using kinematics, we see that the total displacement of the block between the extremes of its motion is− 0. 35 m−
(+ 0. 35 m) =− 0. 70 m. We can safely assume that the time it takes to travel from 0.35m to -0.35m is half the period
T, since other than the direction of the block’s motion, the second half of its cycle is identical to the first half of its
cycle. (The same reasoning would lead us to assume the time from maximum extension to equilibrium must be one-
fourth of the period). The time of travel ist=T 2 =^1.^622 s= 0. 81 s. The average velocity is thenv=∆∆xt=− 00 .. 8170 sm=
− 0. 864 →− 0. 86 ms and the time to travel from +0.35m to equilibrium 0.00m ist=∆vx=−− 00 .. 86435 mm
s
= 0. 405 → 0. 41 s
Check Your Understanding
1a. We know that at the instant the block in the previous problem is released, its instantaneous speed is zero and its
speed at equilibrium is 1.4m/s.
Why can’t we also use the “good old” kinematic equationx=vi+ 2 vftwhere the distancex= 0. 35 m,vi+ 2 vf=^0 +( 21.^4 )=
0. 7 ms and just solve for time? If we did, we would have hadt=^00 ..^3570 mms = 0. 50 s. This result is not in agreement with
the previous result of 0.41 s. Why do you think this is?
Answer:The equationx=vi+ 2 vftis only true if the acceleration is constant. Since the force on the block isF=
−kx→ma=−kx, the accelerationavaries withxand is therefore not constant.
1b. How does the period of oscillation of the block in the previous example change if the spring is stretched twice
as much, that is, if the amplitude of oscillations are doubled to 0.70 m?
The period would be:
a. The same.
b. Half as great.
c. Twice as great.
Answer:The answer is A.
The farther the spring is stretched, the greater the force on the object attached to the spring (recall thatF∝−x→
a∝−x) and so the greater the average speed of the block. The larger average speed makes up exactly for the greater
distance the block must travel during one cycle.
1c. If the spring constant is increased, the period is:
a. The same
b. Decreased
c. Increased
Answer:The answer is B. The spring constant is in the denominator, so a larger value must make the resultT
smaller. This is reasonable, since the larger the spring constant, the stiffer the spring and so the more force pulling
on the object, shortening the time needed to complete a cycle of motion. What effect do you think increasing the
mass would have on the period? Would twice the mass double the period?
We end this section with an example of an object attached to a vertical spring.