Energy.

Figure 1.4:

In a spring-mass system the particle has a potential energy as shown in Figure 1.4. This energy is stored in the spring when it is either compressed or stretched. The potential energy of the system

$$U=\frac{1}{2} k A^2 \cos^2 (\omega_0t + \phi) =\frac{1}{4} m \omega_0^2 A^2 \{ 1 + \cos[ 2 (\omega_0t + \phi)] \}$$ (1.11)

oscillates with angular frequency $2 \omega_0$ as the spring is alternately compressed and stretched. The kinetic energy $m v^2/2$

$$T=\frac{1}{2} m \omega_0^2 A^2 \sin^2 (\omega_0t + \phi) ... ...}{4} m \omega_0^2 A^2 \{ 1 - \cos[ 2 (\omega_0t + \phi)] \}$$ (1.12)

shows similar oscillations which are exactly $\pi$ out of phase.

In a spring-mass system the total energy oscillates between the potential energy of the spring ($U$) and the kinetic energy of the mass ($T$). The total energy $E=T+U$ has a value which remains constant.

The average value of an oscillating quantity is often of interest. We denote the time average of any quantity $Q(t)$ using $\langle Q \rangle$ which is defined as

$$\langle Q \rangle = \lim_{T \rightarrow \infty} \frac{1}{T} \int_{-T/2}^{T/2} Q(t) dt \,.$$ (1.13)

The basic idea here is to average over a time interval $T$ which is significantly larger than the oscillation time period.

It is very useful to remember that $\langle \cos(\omega_0 t + \phi) \rangle =0$. This can be easily verified by noting that the values $\sin(\omega_0 t + \phi)$ are bound between $-1$ and

$+1$. We use this to calculate the average kinetic and potential energies both of which have the same values

$$\langle U \rangle = \langle T \rangle = \frac{1}{4} m \omega_0^2 A^2 \,.$$ (1.14)

The average kinetic and potential energies, and the total energy are all very conveniently expressed in the complex representation as

$$E/2= \langle U \rangle = \langle T \rangle = \frac{1}{4} m \tilde{v} \tilde{v}^{*}=\frac{1}{4} k \tilde{x}\tilde{x}^{*}$$ (1.15)

where ${}^*$ denotes the conjugate of a complex number.
Problem 3: The mean displacement of a SHO $\langle x \rangle$ is zero. The root mean square (rms.) displacement $\sqrt{\langle x^2 \rangle}$ is useful in quantifying the amplitude of oscillation. Verify that the rms. displacement is $\sqrt{\tilde{x} \tilde{x}*/2}$.
Solution: $\sqrt{\langle x^2(t) \rangle}=\sqrt {A^2 \langle \cos^2(\omega_0 t + \phi) \ra... ...^{i \omega } \tilde{A}^{*} e^{- i \omega t}/2}=\sqrt{\tilde{x}\tilde{x}^*/2}$