In this sub-chapter we talk about:

Energy considerations in SHM
Spring potential energy
Energy conversion and graphical representation

Potential and kinetic energy

In simple harmonic motion (SHM), the system’s mechanical energy continually shifts between:

elastic potential energy stored in the spring
kinetic energy of the moving mass

Because we’re assuming no non-conservative forces (a frictionless system), the total mechanical energy stays constant.

Spring potential energy
When a spring with constant $k$ is displaced by $x$ from its equilibrium length, the work required to stretch or compress it becomes stored as spring potential energy:

$U_{s} (x) = \int_{0}^{x} k x^{'} d x^{'} = \frac{1}{2} k x^{2} .$

Kinetic energy
A mass $m$ moving with speed $v$ has kinetic energy:

$K = \frac{1}{2} m v^{2} .$

In SHM, $v$ changes with position $x$ . We can find that relationship using energy conservation.

Total energy and relationship
At any instant, the total mechanical energy is

$E = K + U_{s} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = constant .$

At the turning points $x = \pm A$ , the mass is momentarily at rest ( $v = 0$ ), so all the energy is spring potential energy:

$E = U_{s} (\pm A) = \frac{1}{2} k A^{2} .$

That means for any displacement $x$ ,

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

which we can rearrange to solve for the speed as a function of position:

$v (x) = \pm ω A^{2} - x^{2}, ω = \frac{k}{m} .$

Energy exchange and graphical representation

You can visualize the energy exchange in SHM by plotting kinetic and potential energy versus displacement $x$ or versus time $t$ .

Energy vs. displacement

In this normalized plot ( $A = 1$ , $E = \frac{1}{2} k A^{2} = 1$ ):

$K$ is maximum at $x = 0$ and zero at $x = \pm 1$ .
$U_{s}$ is zero at $x = 0$ and maximum at $x = \pm 1$ .

This matches the physical picture: the mass moves fastest at equilibrium and stops at the turning points.

Energy vs. time

The kinetic and potential energies oscillate out of phase: when one is maximum, the other is zero. Throughout the motion, their sum stays constant at $E = \frac{1}{2} k A^{2}$ .

Dynamics: force and acceleration restoring force and acceleration

The spring’s restoring force points toward equilibrium and is proportional to displacement:

$F = - k x .$

Applying Newton’s second law gives the SHM equation of motion:

$m \overset{x}{¨} = - k x ⟹ \overset{x}{¨} + ω^{2} x = 0, ω^{2} = \frac{k}{m} .$

So the acceleration is proportional to $x$ and always directed toward equilibrium.

Reasoning practice

A block of mass $m$ on a frictionless horizontal spring oscillates with small amplitude $A_{1}$ . It is then given an impulse so that its new amplitude is $A_{2} = 2 A_{1}$ .

(a) How does the period $T$ compare before and after?

Answer: $T = 2 π m / k$ is independent of amplitude, so

$T_{before} = T_{after} .$

(b) How does the total mechanical energy $E$ compare before and after?

Answer:

$E = \frac{1}{2} k A^{2},$

$E_{2} / E_{1} = \frac{\frac{1}{2} k ( 2 A _{1} ) ^{2}}{\frac{1}{2} k A _{1}^{2}} = 4,$

i.e. $E_{2} = 4 E_{1}$ .

Answer:

$\frac{K}{E} = 1 - (\frac{x}{A})^{2}, \frac{U _{s}}{E} = (\frac{x}{A})^{2} .$

Before: $x / A_{1} = 0.5 \Rightarrow K_{1} / E_{1} = 0.75, U_{1} / E_{1} = 0.25$ .

After: $x / A_{2} = 0.25 \Rightarrow K_{2} / E_{2} = 0.9375, U_{2} / E_{2} = 0.0625$ . Since $E_{2} = 4 E_{1}$ ,

$K_{1} = 0.75 E_{1}, U_{1} = 0.25 E_{1}; K_{2} = 0.9375 \times 4 E_{1} = 3.75 E_{1}, U_{2} = 0.0625 \times 4 E_{1} = 0.25 E_{1} .$

Thus $K_{2} > K_{1}$ (five-times larger) while $U_{2} = U_{1}$ .

A simple pendulum of length $L$ swings with small amplitude inside a car. First the car is at rest, then it accelerates forward at constant $a$ .

(a) How does the period compare before and after?

Answer: In the accelerating frame the effective gravity is $g_{eff} = g^{2} + a^{2}$ , so

$T_{new} = 2 π \frac{L}{g _{eff}} < 2 π \frac{L}{g} = T_{old} .$

(b) Explain qualitatively how the restoring torque changes.

Answer: The torque about the pivot is $τ = - m g_{eff} L θ$ for small $θ$ . Since $g_{eff} > g$ , the magnitude of the restoring torque is larger, so the pendulum swings back toward equilibrium more strongly.

Answer:

$U (θ) = m gL (1 - cos θ) ⟶ m g_{eff} L (1 - cos θ) .$

On the same axes the post-acceleration curve is steeper (a narrower well) than the original.

Potential energy before and after acceleration

Two mass-spring systems execute SHM with the same amplitude $A$ and period $T$ . In system I: $(m_{I}, k_{I})$ ; in II: $(m_{II} = 4 m_{I}, k_{II} = 4 k_{I})$ .

(a) Verify both have the same period.

Answer:

$T = 2 π \frac{m}{k}, T_{I} = 2 π \frac{m _{I}}{k _{I}}, T_{II} = 2 π \frac{4 m _{I}}{4 k _{I}} = T_{I} .$

(b) At $x = A /2$ , compare the speeds.

Answer:

$v = ω A^{2} - x^{2}, ω = \frac{k}{m} .$

Since $k / m$ is the same in both, and $A$ is the same, $v_{I} = v_{II}$ .

Answer: $F_{m a x} = k A$ , so

$F_{m a x, I} = k_{I} A, F_{m a x, II} = 4 k_{I} A = 4 F_{m a x, I} .$

Thus system II’s spring exerts four times the maximum force.

Solved examples

Example Problem 1 : (exam style FRQ)
A block of mass $m$ is attached to a horizontal spring of force constant $k$ on a frictionless surface. The block is pulled to the right until the spring is stretched by an amount $A$ and released from rest. The block then executes simple harmonic motion about the equilibrium position.

Spring mass system according to the question

(a) Derive an expression for the total mechanical energy $E$ of the oscillating block in terms of $k$ and $A$ .

(b) Starting from $E = K + U_{s}$ , show that the speed $v$ of the block as a function of its instantaneous displacement $x$ from equilibrium is

$v (x) = \pm ω A^{2} - x^{2}, ω = \frac{k}{m} .$

(i) Calculate the numerical value of the total energy $E$ (in joules). (ii) Determine the kinetic energy $K$ and spring potential energy $U_{s}$ when $x = 0.060$ m. (iii) Compute the block’s speed $v$ at $x = 0.060$ m.

(d) At what displacement $x$ (in terms of $A$ ) is the kinetic energy equal to one-quarter of the total energy? Show your reasoning.

Solution

(a) At the turning point ( $x = \pm A$ ) all energy is spring potential:

$E = U_{s} (A) = \frac{1}{2} k A^{2} .$

(b) Start with

$E = K + U_{s} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2} .$

Rearrange for $v^{2}$ :

$\frac{1}{2} m v^{2} = \frac{1}{2} k (A^{2} - x^{2}) ⟹ v^{2} = \frac{k}{m} (A^{2} - x^{2}) = ω^{2} (A^{2} - x^{2}) .$

Thus

$v (x) = \pm ω A^{2} - x^{2} .$

(c)

(i)

$E = \frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.100)^{2} = 1.00 J .$

(ii) At $x = 0.060$ m,

$U_{s} = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.060)^{2} = 0.36 J,$

$K = E - U_{s} = 1.00 - 0.36 = 0.64 J .$

(iii)

$v = \frac{2 K}{m} = \frac{2 ( 0.64 )}{0.500} = 2.56 = 1.60 m/s .$

(d) We require

$K = \frac{1}{4} E ⟹ 1 - \frac{x ^{2}}{A ^{2}} = \frac{1}{4} ⟹ \frac{x ^{2}}{A ^{2}} = \frac{3}{4} ⟹ x = \pm \frac{3}{2} A .$

Example Problem 2:
A block $m = 0.300$ kg on a frictionless table is attached to a spring with $k = 80.0$ N/m. It is pulled out to $A = 0.150$ m and released from rest.

(a) Compute the total energy $E$ . (b) At what displacement $x$ is the kinetic energy equal to the potential energy? (c) What is the speed at that point? (d) What fraction of the total energy is kinetic at $x = 0.100$ m?

Solution.

(a)

$E = \frac{1}{2} k A^{2} = \frac{1}{2} (80.0) (0.150)^{2} = 0.900 J .$

(b) Set $K = U_{s} = \frac{1}{2} E = 0.450$ J:

$\frac{1}{2} k x^{2} = 0.450 ⟹ x^{2} = \frac{2 ( 0.450 )}{80.0} = 0.01125 ⟹ x = 0.106 m .$

(c)

$K = 0.450 J = \frac{1}{2} m v^{2} ⟹ v = \frac{2 ( 0.450 )}{0.300} = 3.00 = 1.732 m/s .$

(d) At $x = 0.100$ m,

$U_{s} = \frac{1}{2} (80.0) (0.100)^{2} = 0.400 J, K = 0.900 - 0.400 = 0.500 J,$

fraction kinetic = $0.500/0.900 \approx 0.556$ .

Example Problem 3:
A simple pendulum of length $L = 0.800$ m oscillates with small amplitude $Θ = 0.200$ rad.

(a) Find its total energy in terms of mass $m$ . (b) When the angular displacement is $θ = 0.150$ rad, find the potential and kinetic energies. (c) Determine the angular speed $ω$ at $θ = 0.150$ rad.

Solution.

(a)

$E = \frac{1}{2} m gL Θ^{2} = \frac{1}{2} m (9.80) (0.800) (0.200)^{2} = 0.1568 m J .$

(b) At $θ = 0.150$ rad,

$U = \frac{1}{2} m gL θ^{2} = 0.0882 m J, K = 0.1568 m - 0.0882 m = 0.0686 m J .$

$ω = \frac{2 ( 0.0686 m )}{m ( 0.800 ) ^{2}} = 0.463 rad/s .$

In this sub-chapter we talk about:

Energy considerations in SHM
Spring potential energy
Energy conversion and graphical representation

Potential and kinetic energy

In simple harmonic motion (SHM), the system’s mechanical energy continually shifts between:

elastic potential energy stored in the spring
kinetic energy of the moving mass

Because we’re assuming no non-conservative forces (a frictionless system), the total mechanical energy stays constant.

Spring potential energy
When a spring with constant $k$ is displaced by $x$ from its equilibrium length, the work required to stretch or compress it becomes stored as spring potential energy:

$U_{s} (x) = \int_{0}^{x} k x^{'} d x^{'} = \frac{1}{2} k x^{2} .$

Kinetic energy
A mass $m$ moving with speed $v$ has kinetic energy:

$K = \frac{1}{2} m v^{2} .$

In SHM, $v$ changes with position $x$ . We can find that relationship using energy conservation.

Total energy and relationship
At any instant, the total mechanical energy is

$E = K + U_{s} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = constant .$

At the turning points $x = \pm A$ , the mass is momentarily at rest ( $v = 0$ ), so all the energy is spring potential energy:

$E = U_{s} (\pm A) = \frac{1}{2} k A^{2} .$

That means for any displacement $x$ ,

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

which we can rearrange to solve for the speed as a function of position:

$v (x) = \pm ω A^{2} - x^{2}, ω = \frac{k}{m} .$

Energy exchange and graphical representation

You can visualize the energy exchange in SHM by plotting kinetic and potential energy versus displacement $x$ or versus time $t$ .

Energy vs. displacement

In this normalized plot ( $A = 1$ , $E = \frac{1}{2} k A^{2} = 1$ ):

$K$ is maximum at $x = 0$ and zero at $x = \pm 1$ .
$U_{s}$ is zero at $x = 0$ and maximum at $x = \pm 1$ .

This matches the physical picture: the mass moves fastest at equilibrium and stops at the turning points.

Energy vs. time

The kinetic and potential energies oscillate out of phase: when one is maximum, the other is zero. Throughout the motion, their sum stays constant at $E = \frac{1}{2} k A^{2}$ .

Dynamics: force and acceleration restoring force and acceleration

The spring’s restoring force points toward equilibrium and is proportional to displacement:

$F = - k x .$

Applying Newton’s second law gives the SHM equation of motion:

$m \overset{x}{¨} = - k x ⟹ \overset{x}{¨} + ω^{2} x = 0, ω^{2} = \frac{k}{m} .$

So the acceleration is proportional to $x$ and always directed toward equilibrium.

Reasoning practice

A block of mass $m$ on a frictionless horizontal spring oscillates with small amplitude $A_{1}$ . It is then given an impulse so that its new amplitude is $A_{2} = 2 A_{1}$ .

(a) How does the period $T$ compare before and after?

Answer: $T = 2 π m / k$ is independent of amplitude, so

$T_{before} = T_{after} .$

(b) How does the total mechanical energy $E$ compare before and after?

Answer:

$E = \frac{1}{2} k A^{2},$

$E_{2} / E_{1} = \frac{\frac{1}{2} k ( 2 A _{1} ) ^{2}}{\frac{1}{2} k A _{1}^{2}} = 4,$

i.e. $E_{2} = 4 E_{1}$ .

Answer:

$\frac{K}{E} = 1 - (\frac{x}{A})^{2}, \frac{U _{s}}{E} = (\frac{x}{A})^{2} .$

Before: $x / A_{1} = 0.5 \Rightarrow K_{1} / E_{1} = 0.75, U_{1} / E_{1} = 0.25$ .

After: $x / A_{2} = 0.25 \Rightarrow K_{2} / E_{2} = 0.9375, U_{2} / E_{2} = 0.0625$ . Since $E_{2} = 4 E_{1}$ ,

$K_{1} = 0.75 E_{1}, U_{1} = 0.25 E_{1}; K_{2} = 0.9375 \times 4 E_{1} = 3.75 E_{1}, U_{2} = 0.0625 \times 4 E_{1} = 0.25 E_{1} .$

Thus $K_{2} > K_{1}$ (five-times larger) while $U_{2} = U_{1}$ .

A simple pendulum of length $L$ swings with small amplitude inside a car. First the car is at rest, then it accelerates forward at constant $a$ .

(a) How does the period compare before and after?

Answer: In the accelerating frame the effective gravity is $g_{eff} = g^{2} + a^{2}$ , so

$T_{new} = 2 π \frac{L}{g _{eff}} < 2 π \frac{L}{g} = T_{old} .$

(b) Explain qualitatively how the restoring torque changes.

Answer:

$U (θ) = m gL (1 - cos θ) ⟶ m g_{eff} L (1 - cos θ) .$

On the same axes the post-acceleration curve is steeper (a narrower well) than the original.

Two mass-spring systems execute SHM with the same amplitude $A$ and period $T$ . In system I: $(m_{I}, k_{I})$ ; in II: $(m_{II} = 4 m_{I}, k_{II} = 4 k_{I})$ .

(a) Verify both have the same period.

Answer:

$T = 2 π \frac{m}{k}, T_{I} = 2 π \frac{m _{I}}{k _{I}}, T_{II} = 2 π \frac{4 m _{I}}{4 k _{I}} = T_{I} .$

(b) At $x = A /2$ , compare the speeds.

Answer:

$v = ω A^{2} - x^{2}, ω = \frac{k}{m} .$

Since $k / m$ is the same in both, and $A$ is the same, $v_{I} = v_{II}$ .

Answer: $F_{m a x} = k A$ , so

$F_{m a x, I} = k_{I} A, F_{m a x, II} = 4 k_{I} A = 4 F_{m a x, I} .$

Thus system II’s spring exerts four times the maximum force.

Solved examples

(a) Derive an expression for the total mechanical energy $E$ of the oscillating block in terms of $k$ and $A$ .

(b) Starting from $E = K + U_{s}$ , show that the speed $v$ of the block as a function of its instantaneous displacement $x$ from equilibrium is

$v (x) = \pm ω A^{2} - x^{2}, ω = \frac{k}{m} .$

(d) At what displacement $x$ (in terms of $A$ ) is the kinetic energy equal to one-quarter of the total energy? Show your reasoning.

Solution

(a) At the turning point ( $x = \pm A$ ) all energy is spring potential:

$E = U_{s} (A) = \frac{1}{2} k A^{2} .$

(b) Start with

$E = K + U_{s} = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2} = \frac{1}{2} k A^{2} .$

Rearrange for $v^{2}$ :

$\frac{1}{2} m v^{2} = \frac{1}{2} k (A^{2} - x^{2}) ⟹ v^{2} = \frac{k}{m} (A^{2} - x^{2}) = ω^{2} (A^{2} - x^{2}) .$

Thus

$v (x) = \pm ω A^{2} - x^{2} .$

(c)

(i)

$E = \frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.100)^{2} = 1.00 J .$

(ii) At $x = 0.060$ m,

$U_{s} = \frac{1}{2} k x^{2} = \frac{1}{2} (200) (0.060)^{2} = 0.36 J,$

$K = E - U_{s} = 1.00 - 0.36 = 0.64 J .$

(iii)

$v = \frac{2 K}{m} = \frac{2 ( 0.64 )}{0.500} = 2.56 = 1.60 m/s .$

(d) We require

$K = \frac{1}{4} E ⟹ 1 - \frac{x ^{2}}{A ^{2}} = \frac{1}{4} ⟹ \frac{x ^{2}}{A ^{2}} = \frac{3}{4} ⟹ x = \pm \frac{3}{2} A .$

Example Problem 2:
A block $m = 0.300$ kg on a frictionless table is attached to a spring with $k = 80.0$ N/m. It is pulled out to $A = 0.150$ m and released from rest.

Solution.

(a)

$E = \frac{1}{2} k A^{2} = \frac{1}{2} (80.0) (0.150)^{2} = 0.900 J .$

(b) Set $K = U_{s} = \frac{1}{2} E = 0.450$ J:

$\frac{1}{2} k x^{2} = 0.450 ⟹ x^{2} = \frac{2 ( 0.450 )}{80.0} = 0.01125 ⟹ x = 0.106 m .$

(c)

$K = 0.450 J = \frac{1}{2} m v^{2} ⟹ v = \frac{2 ( 0.450 )}{0.300} = 3.00 = 1.732 m/s .$

(d) At $x = 0.100$ m,

$U_{s} = \frac{1}{2} (80.0) (0.100)^{2} = 0.400 J, K = 0.900 - 0.400 = 0.500 J,$

fraction kinetic = $0.500/0.900 \approx 0.556$ .

Example Problem 3:
A simple pendulum of length $L = 0.800$ m oscillates with small amplitude $Θ = 0.200$ rad.

Solution.

(a)

$E = \frac{1}{2} m gL Θ^{2} = \frac{1}{2} m (9.80) (0.800) (0.200)^{2} = 0.1568 m J .$

(b) At $θ = 0.150$ rad,

$U = \frac{1}{2} m gL θ^{2} = 0.0882 m J, K = 0.1568 m - 0.0882 m = 0.0686 m J .$

$ω = \frac{2 ( 0.0686 m )}{m ( 0.800 ) ^{2}} = 0.463 rad/s .$

Energy and dynamics in SHM

Potential and kinetic energy

Energy exchange and graphical representation

Dynamics: force and acceleration restoring force and acceleration

Reasoning practice

Solved examples

Energy and dynamics in SHM

Potential and kinetic energy

Energy exchange and graphical representation

Dynamics: force and acceleration restoring force and acceleration

Reasoning practice

Solved examples