15.3.1 Conservation of Energy

Because energy is conserved, we can combine Equations 15.18, 15.19, and 15.20 to write

E = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2} = \frac{1}{2} m (v_{max})^{2} (conservation of energy)

(15.25)

Any pair of these expressions may be useful, depending on the known information. For example, you can use the amplitude A to find the speed at any point x by combining the first and second expressions for E. The speed v at position x is

v = \sqrt{\frac{k}{m} (A^{2} - x^{2})} = ω \sqrt{A^{2} - x^{2}}

(15.26)

FIGURE 15.9 shows graphically how the kinetic and potential energy change with time. They both oscillate but remain positive because x and $v$ are squared. Energy is continuously being transformed back and forth between the kinetic energy of the moving block and the stored potential energy of the spring, but their sum remains constant. Notice that K and U both oscillate twice each period; make sure you understand why.

The figure illustrates an energy versus time curve and a position versus time curve.

FIGURE 15.9 Full Alternative Text

EXAMPLE 15.4 Using conservation of energy

A 500 g block on a spring is pulled a distance of 20 cm and released. The subsequent oscillations are measured to have a period of 0.80 s.

At what position or positions is the block’s speed $1.0 m / s ?$
What is the spring constant?

MODEL The motion is SHM. Energy is conserved.

SOLVE a. The block starts from the point of maximum displacement, where $E = U = \frac{1}{2} k A^{2}$ . At a later time, when the position is x and the speed is $v$ , energy conservation requires

\frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2}

Solving for x, we find

x = \sqrt{A^{2} - \frac{m v^{2}}{k}} = \sqrt{A^{2} - {(\frac{v}{ω})}^{2}}

where we used $k / m = ω^{2}$ from Equations 15.24. The angular frequency is easily found from the period: $ω = 2 π / T = 7.85 rad / s$ . Thus

x = \sqrt{(0.20 {m)}^{2} - (\frac{1.0 m/s}{7.85 rad/s})^{2}} = \pm 0.15 m = \pm 15 cm

There are two positions because the block has this speed on either side of equilibrium.

b. Although part a did not require that we know the spring constant, it is straightforward to find from Equations 15.24:

\begin{array}{l} T = 2 π \sqrt{\frac{m}{k}} \\ k = \frac{4 π^{2} m}{T^{2}} = \frac{4 π^{2} (0.50 kg)}{{(0.80 s)}^{2}} = 31 N / m \end{array}