15-2 Energy in Simple Harmonic Motion

15-2 Energy in Simple Harmonic Motion#

Prompts

  • For a spring–block oscillator, where is the energy stored? How does kinetic energy \(K\) and potential energy \(U\) change as the block moves?

  • Why is mechanical energy \(E = K + U\) constant in undamped SHM? At what positions is \(E\) entirely kinetic? Entirely potential?

  • Sketch how \(K(t)\), \(U(t)\), and \(E\) look versus time. How many times per period does each of \(K\) and \(U\) peak?

  • Given amplitude \(x_m\), spring constant \(k\) and block mass \(m\), how do you find the maximum speed \(v_m\) of the block? Use energy conservation.

Lecture Notes#

Overview#

  • In a spring–block oscillator, energy shuttles between kinetic (block) and elastic potential (spring); the mechanical energy \(E = K + U\) is conserved (no friction).

  • Kinetic energy \(K = \frac{1}{2}mv^2\) peaks when the block passes through equilibrium (\(x = 0\)); potential energy \(U = \frac{1}{2}kx^2\) peaks at the extremes (\(x = \pm x_m\)).

  • Energy conservation gives a direct path to relate amplitude, maximum speed, and spring constant—often simpler than using \(x(t)\) and \(v(t)\).

Location

Kinetic \(K\)

Potential \(U\)

\(x = 0\) (center)

Maximum

Zero

\(x = \pm x_m\) (extremes)

Zero

Maximum

Kinetic and potential energy#

For a spring–block oscillator (see section 15-1):

  • Potential energy (stored in the spring):

(53)#\[ U = \frac{1}{2}kx^2 \]

  • Kinetic energy (stored in the block):

(54)#\[ K = \frac{1}{2}mv^2 \]

Both \(U\) and \(K\) vary with time as the block oscillates. At any instant, their sum is the mechanical energy \(E\).

Two elements of any oscillator

Every oscillating system needs

  1. inertia—to store kinetic energy,

  2. stiffness—to store potential energy.

In the spring–block model, these are separated: the block holds \(K\) and the spring holds \(U\).

Conservation of mechanical energy#

With no friction, \(E = K + U\) is constant:

(55)#\[ E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \text{constant} \]

The simplest way to evaluate \(E\) is at an extreme, where \(v = 0\) and \(x = x_m\):

(56)#\[ E = \frac{1}{2}kx_m^2 \]

Alternatively, at the center (\(x = 0\)) where speed is maximum (\(v = v_m\)):

(57)#\[ E = \frac{1}{2}mv_m^2 \]

Equating the two expressions links amplitude and maximum speed:

(58)#\[ \frac{1}{2}kx_m^2 = \frac{1}{2}mv_m^2 \quad \Rightarrow \quad v_m = \omega x_m \]

with \(\omega = \sqrt{k/m}\).

Reading off period from energy

Every harmonic oscillator has energy in the general form

\[ E = \frac{1}{2}\mu\,\left(\frac{dq}{dt}\right)^2 + \frac{1}{2}\kappa\,q^2 \]

Symbol

Name

Meaning

\(q\)

Generalized coordinate

The oscillating quantity (e.g., position \(x\) for spring–block)

\(\mu\)

Generalized inertia

Coefficient of the kinetic term (e.g., mass \(m\) for spring–block)

\(\kappa\)

Generalized stiffness

Coefficient of the potential term (e.g., spring constant \(k\) for spring–block)

From this form, the angular frequency and period follow directly:

(59)#\[ \omega = \sqrt{\frac{\kappa}{\mu}}, \qquad T = 2\pi\sqrt{\frac{\mu}{\kappa}} \]

This result applies to all SHM—spring–block, torsion pendulum, pendulums, and beyond.

Why the read-off works. Energy conservation \(dE/dt = 0\) gives

\[ \left(\mu\,\frac{d^2q}{dt^2} + \kappa\,q\right)\frac{dq}{dt} = 0 \]

Dividing by \(dq/dt\) (nonzero except at turning points) yields \(\mu\,d^2q/dt^2 = -\kappa q\). This is exactly the SHM hallmark from section 15-1:

\[ \frac{d^2q}{dt^2} = -\omega^2 q \]

with \(\omega^2 = \kappa/\mu\). Once the energy has the quadratic form, \(\mu\) and \(\kappa\) determine \(\omega\)—no separate force analysis is needed.

Energy as a function of position and time#

Versus position \(x\): \(U(x) = \frac{1}{2}kx^2\) is a parabola; \(K(x) = E - U(x)\) is the complement. At \(x = 0\) the energy is all kinetic; at \(x = \pm x_m\) it is all potential.

Versus time \(t\): \(U(t)\) and \(K(t)\) oscillate—each peaks twice per period (at \(t\) and \(t + T/2\)). Their sum \(E\) stays constant.

Verification: \(E(t)\) independent of \(t\)

With \(x(t) = x_m \cos(\omega t + \phi)\) and \(v(t) = -\omega x_m \sin(\omega t + \phi)\) (from section 15-1):

(60)#\[ U(t) = \frac{1}{2}kx^2(t) = \frac{1}{2}k x_m^2 \cos^2(\omega t + \phi) \]
(61)#\[ K(t) = \frac{1}{2}mv^2(t) = \frac{1}{2}m \omega^2 x_m^2 \sin^2(\omega t + \phi) \]

For the spring–block oscillator, \(\omega = \sqrt{k/m}\), so \(m\omega^2 = k\). Substituting into Eq. (61):

(62)#\[ K(t) = \frac{1}{2}k x_m^2 \sin^2(\omega t + \phi) \]

Adding Eq. (60) and Eq. (62):

(63)#\[ E(t) = K(t) + U(t) = \frac{1}{2}k x_m^2 \left[\cos^2(\omega t + \phi) + \sin^2(\omega t + \phi)\right] = \frac{1}{2}k x_m^2 \]

Since \(\cos^2\theta + \sin^2\theta = 1\), the bracket equals 1; \(E(t)\) is constant—independent of \(t\).

Poll: Energy at different positions

A block has kinetic energy 3 J and potential energy 2 J when at \(x = +2\;\text{cm}\). What is the kinetic energy when the block is at \(x = 0\)?

  • (A) 0 J

  • (B) 2 J

  • (C) 3 J

  • (D) 5 J

Summary#

  • \(U = \frac{1}{2}kx^2\), \(K = \frac{1}{2}mv^2\); mechanical energy \(E = K + U\) is constant in undamped SHM.

  • \(E = \frac{1}{2}kx_m^2 = \frac{1}{2}mv_m^2\)—evaluate at extremes or center to relate amplitude and max speed.

  • \(v_m = \omega x_m\) from energy conservation.

  • At \(x = 0\): all kinetic. At \(x = \pm x_m\): all potential.