SAT Physics Subject Test
Chapter 7 Oscillations
THE SPRING–BLOCK OSCILLATOR: VERTICAL MOTION
So far we”ve looked at a block sliding back and forth on a horizontal table, but the block could also oscillate vertically. The only difference would be that gravity would cause the block to move downward, to an equilibrium position at which, in contrast with the horizontal SHM we”ve examined, the spring would not be at its natural length.
Vertical vs. Horizontal
Use what you know about
horizontal motion of the
spring–block oscillator.
The only difference in
vertical motion is that you
must account for gravity.
Consider a spring of negligible mass hanging from a stationary support. A block of mass m is attached to its end and allowed to come to rest, stretching the spring a distance d. At this point, the block is in equilibrium; the upward force of the spring is balanced by the downward force of gravity.
Therefore
Next, imagine that the block is pulled down a distance A and released. The spring force increases (because the spring was stretched farther); it”s stronger than the block”s weight, and, as a result, the block accelerates upward. As the block”s momentum carries it up, through the equilibrium position, the spring becomes less stretched than it was at equilibrium, so FS is less than the block”s weight. As a result, the block decelerates, stops, and accelerates downward again, and the up-and-down motion repeats.
When the block is at a distance y below its equilibrium position, the spring is stretched a total distance of d + y, so the upward spring force is equal to k(d + y), while the downward force stays the same, mg.
The net force on the block is
F = k(d + y) – mg
but this equation becomes F = ky because kd = mg (as we saw above).
Since the resulting force on the block, F = ky, has the form of Hooke”s law, we know that the vertical simple harmonic oscillations of the block have the same characteristics as do horizontal oscillations, with the equilibrium position, y = 0, not at the spring”s natural length, but at the point where the hanging block is in equilibrium.
Questions 10-13
A block of mass m = 1.5 kg is attached to the end of a vertical spring of force constant k = 300 N/m. After the block comes to rest, it is pulled down a distance of 2.0 cm and released.
10. How far does the weight of the block cause the spring to stretch initially?
11. What are the minimum and maximum amounts of stretch of the spring during the oscillations of the block?
12. At what point(s) will the speed of the block be zero?
13. At what point(s) will the acceleration of the block be zero?
Here”s How to Crack It
10. The weight of the block initially stretches the spring by a distance of
11. Since the amplitude of the motion is 2.0 cm, the spring is stretched a maximum of 5 cm + 2.0 cm = 7 cm when the block is at the lowest position in its cycle, and a minimum of 5 cm – 2.0 cm = 3 cm when the block is at its highest position.
12. The block”s speed is 0 at the two ends of its oscillation region, which are the points described in question 12.
13. The block”s acceleration is 0 at its equilibrium position, which is the point described in question 11.
