We have focused our analysis on a horizontally oscillating spring. But the typical demonstration you’ll see in class is a mass bobbing up and down on a spring hung vertically from a support. Is it safe to assume that a vertical oscillation has the same mathematical description as a horizontal oscillation? Or does the additional force of gravity change the motion? Let us look at this more carefully.

FIGURE 15.14 shows a block of mass m hanging from a spring of spring constant k. An important fact to notice is that the equilibrium position of the block is not where the spring is at its unstretched length. At the equilibrium position of the block, where it hangs motionless, the spring has stretched by $\text{Δ}L$.

FIGURE 15.14 Gravity stretches the spring.

Finding $\text{Δ}L$ is an equilibrium problem in which the upward spring force balances the downward gravitational force on the block. The y-component of the spring force is given by Hooke’s law:

Equation 15.38 makes a distinction between $\text{Δ}L$, which is simply a distance and is a positive number, and the displacement $\text{Δ}y$. The block is displaced downward, so $\text{Δ}y=-\text{Δ}L$. Newton’s first law for the block in equilibrium is

from which we can find

This is the distance the spring stretches when the block is attached to it.

Let the block oscillate around this equilibrium position, as shown in FIGURE 15.15. We’ve now placed the origin of the y-axis at the block’s equilibrium position in order to be consistent with our analyses of oscillations throughout this chapter. If the block moves upward, as the figure shows, the spring gets shorter compared to its equilibrium length, but the spring is still stretched compared to its unstretched length in Figure 15.14. When the block is at position y, the spring is stretched by an amount $\text{Δ}L-y$ and hence exerts an upward spring force $F_{\text{Sp}}=k(\text{Δ}L-y)$. The net force on the block at this point is

But $k\text{Δ}L-mg$ is zero, from Equation 15.40, so the net force on the block is simply

FIGURE 15.15 The block oscillates around the equilibrium position.

Equation 15.42 for vertical oscillations is exactly the same as Equation 15.30 for horizontal oscillations, where we found $(F_{\text{net}}){}_x=-kx$. That is, the restoring force for vertical oscillations is identical to the restoring force for horizontal oscillations. The role of gravity is to determine where the equilibrium position is, but it doesn’t affect the oscillatory motion around the equilibrium position.

Because the net force is the same, Newton’s second law has exactly the same oscillatory solution:

with, again, $\omega =\sqrt{k\text{/}m}$. The vertical oscillations of a mass on a spring are the same simple harmonic motion as those of a block on a horizontal spring. This is an important finding because it was not obvious that the motion would still be simple harmonic motion when gravity was included.

EXAMPLE 15.6 Bungee oscillations

An 83 kg student hangs from a bungee cord with spring constant $270\text{N}\text{/}\text{m}$. The student is pulled down to a point where the cord is 5.0 m longer than its unstretched length, then released. Where is the student, and what is his velocity 2.0 s later?

MODEL A bungee cord can be modeled as a spring. Vertical oscillations on the bungee cord are SHM.

VISUALIZE FIGURE 15.16 shows the situation.

FIGURE 15.16 A student on a bungee cord oscillates about the equilibrium position.

FIGURE 15.16 Full Alternative Text

SOLVE Although the cord is stretched by 5.0 m when the student is released, this is not the amplitude of the oscillation. Oscillations occur around the equilibrium position, so we have to begin by finding the equilibrium point where the student hangs motionless. The cord stretch at equilibrium is given by Equation 15.40:

$$\text{Δ}L={\displaystyle \frac{mg}{k}}=3.0\text{m}$$

Stretching the cord 5.0 m pulls the student 2.0 m below the equilibrium point, so $A=2.0\text{m}$. That is, the student oscillates with amplitude $A=2.0\text{m}$ about a point 3.0 m beneath the bungee cord’s original end point. The student’s position as a function of time, as measured from the equilibrium position, is

$$y(t)=(2.0\text{m})\text{cos}(\omega t+{\varphi }_0)$$

where $\omega =\sqrt{k\text{/}m}=1.80\text{rad}\text{/}\text{s}$.

The initial condition

$$y_0=A\text{cos}{\varphi }_0=-A$$

requires the phase constant to be ${\varphi }_0=\pi \text{rad}$. At $t=2.0\text{s}$ the student’s position and velocity are

$$\begin{array}{l}y=(2.0\text{ m})\text{ cos }\text{}\text{}\text{}\left(\text{}(1.80\text{ rad}\text{/}\text{s})(2.0\text{ s})+\pi \text{ rad}\right)=1.8\text{ m }\\ v_y=-\omega A\text{ sin}\left(\omega t+{\varphi }_0\right)=-1.6\text{ m}\text{/}\text{s}\end{array}$$

The student is 1.8 m above the equilibrium position, or 1.2 m below the original end of the cord. Because his velocity is negative, he’s passed through the highest point and is heading down.