Worked example 11.4: Energy in simple harmonic motion

Question: A block of mass $m=4 {\rm kg}$ is attached to a spring, and undergoes simple harmonic motion with a period of $T=0.35 {\rm s}$. The total energy of the system is $E=2.5 {\rm J}$. What is the force constant of the spring? What is the amplitude of the motion?

Answer: The angular frequency of the motion is

$$\omega = \frac{2 \pi}{T} = \frac{2 \pi}{0.35} = 17.95 {\rm rad./s}.$$

Now, $\omega=\sqrt{k/m}$ for a mass on a spring. Rearrangement of this formula yields

$$k = m \omega^2 = 4 \times 17.95\times 17.95 = 1289.1 {\rm N/m}.$$

The total energy of a system executing simple harmonic motion is $E= a^2 k/2$. Rearrangement of this formula gives

$$a = \sqrt{\frac{2 E}{k}} = \sqrt{\frac{2\times 2.5}{1289.1}} = 0.06228 {\rm m}.$$

Thus, the force constant is $1289.1 {\rm N/m}$ and the amplitude is $0.06228 {\rm m}$.