Energy in SHM — KE + PE = ½kA² (constant): Question bank

Two symbols recur, so lock them in first.

True or false — justify

True or false: If you double the amplitude

A

, the total energy

E

also doubles.

False.

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

, so

E

scales with

A^{2}

— double

A

gives

4 \times

the energy, not

2 \times

True or false: At the turning points the kinetic energy is zero.

True. At

x = \pm A

the mass reverses direction, so

v = 0

for an instant, hence

K = \frac{1}{2} m v^{2} = 0

; all the energy sits in

U

True or false: At equilibrium (

x = 0

) the potential energy is zero.

True.

U = \frac{1}{2} k x^{2} = 0

when

x = 0

, so the entire

E

is kinetic — this is where the mass is fastest.

True or false: The kinetic energy oscillates at the same frequency as the displacement

x (t)

False.

K \propto sin^{2} (ω t)

, and

sin^{2}

repeats twice as often as

sin

, so

K

(and

U

) oscillate at

2 ω

— double the frequency of

x (t)

True or false: Over one full cycle the average KE equals the average PE.

True. Each averages to

\frac{1}{4} k A^{2} = \frac{1}{2} E

; energy is shared 50–50 on average even though at any instant the split varies.

True or false: The total energy

E

depends on the mass

m

when written as

\frac{1}{2} k A^{2}

False in that form —

\frac{1}{2} k A^{2}

has no

m

. Mass hides inside

ω

: the equivalent form

E = \frac{1}{2} m ω^{2} A^{2}

shows that with

ω = k / m

the

m

cancels back to

\frac{1}{2} k A^{2}

True or false: A heavier mass on the same spring, pulled to the same amplitude, stores the same total energy.

True.

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

uses only

k

and

A

; the mass changes the speed and period but not the stored energy.

True or false: Halving

x

halves the potential energy.

False.

U = \frac{1}{2} k x^{2} \propto x^{2}

, so halving

x

gives

\frac{1}{4}

of the PE, not half.

True or false: In an ideal (undamped) oscillator,

E

slowly decreases over many cycles.

False. Ideal SHM has only the conservative restoring force and no friction, so

E

is exactly constant. Energy decay happens in Damped Oscillations, not here.

Spot the error

"Velocity is largest at the ends because the spring force is biggest there." — find the flaw.

Big force means big acceleration, not big speed. At

x = \pm A

the mass is momentarily stopped; max speed is at

x = 0

where the force is zero.

"Use

U = m g h

for the spring's stored energy." — find the flaw.

m g h

is gravitational PE. A spring's elastic PE is

\frac{1}{2} k x^{2}

, coming from integrating the restoring force

F = - k x

(see Spring constant & Hooke's Law $F=-kx$).

"Since

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

and

U = \frac{1}{2} k x^{2}

both grow, total energy must grow as the mass speeds up." — find the flaw.

They trade off, they don't both grow together. As

K

rises

U

falls by exactly the same amount, keeping

K + U = E

fixed.

"KE

= \frac{1}{2} k A^{2}

everywhere because that's the total energy." — find the flaw.

\frac{1}{2} k A^{2}

is the total

E

, equal to

K

only at

x = 0

. In general

K = \frac{1}{2} k (A^{2} - x^{2})

, which shrinks as

x

grows toward

A

"At

x = A / 2

the mass is halfway to the wall, so KE is half of maximum." — find the flaw.

x = A / 2

K = \frac{1}{2} k (A^{2} - \frac{A ^{2}}{2}) = \frac{1}{4} k A^{2} = \frac{1}{2} E

. That's where KE equals PE (each half of

E

), not "half of maximum KE" — max KE is

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

, so this is half of max KE, but the reasoning "halfway to the wall" is wrong; it's the equal-split point, and

A / 2 \approx 0.707 A

, not

0.5 A

"Doubling the spring constant

k

doubles the max speed." — find the flaw.

Max speed is

v_{m a x} = ω A = k / m A

, so

v_{m a x} \propto k

. Doubling

k

multiplies

v_{m a x}

2

, not by

2

Why questions

Why is the total energy fixed entirely by the amplitude and not by where the mass currently is?

Because

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

is evaluated at a turning point where

K = 0

and all energy is

\frac{1}{2} k A^{2}

; that value is conserved for the whole motion, so wherever the mass is, the split changes but the sum stays

\frac{1}{2} k A^{2}

Why do KE and PE oscillate at twice the frequency of the displacement?

Both depend on squares (

sin^{2}, cos^{2}

), and squaring folds the negative half of a wave onto the positive, so the pattern repeats twice per displacement cycle — using

sin^{2} θ = \frac{1 - c o s 2 θ}{2}

shows the explicit

2 ω

Why does the mass move fastest at the point where the force on it is zero?

The force is zero at

x = 0

, but the mass has been accelerating the entire trip inward, so it arrives at equilibrium with all energy converted to

K

— maximum speed exactly where the net force momentarily vanishes.

Why does the

A^{2}

scaling of energy reappear when we talk about waves?

A wave is many oscillators each carrying

\frac{1}{2} k A^{2}

, so the transported energy (intensity) also scales as amplitude squared — see Energy in Waves — intensity ∝ amplitude².

Why can we replace

m ω^{2}

with

k

when comparing KE and PE?

Because

ω = k / m

means

ω^{2} = k / m

, so

m ω^{2} = k

; this rewrites KE in terms of

k

and

A

, letting it be compared directly with

U = \frac{1}{2} k x^{2}

in the same units.

Why is the restoring force being conservative the key to constant energy?

A conservative force stores work as PE and returns it exactly with no loss, so no energy leaks out as heat — that's precisely the condition Conservation of Mechanical Energy requires.

Edge cases

What happens to the energy split at the exact instant

x = 0

All energy is kinetic:

K = E = \frac{1}{2} k A^{2}

and

U = 0

. This is the single point of maximum speed.

What is the KE at

x = A

(a turning point)?

Zero.

K = \frac{1}{2} k (A^{2} - A^{2}) = 0

; the mass is instantaneously at rest with all energy in

U = \frac{1}{2} k A^{2}

Degenerate case: if the amplitude

A = 0

, what is the motion and the energy?

There is no motion — the mass sits at equilibrium forever.

E = \frac{1}{2} k (0)^{2} = 0

: both banks are empty.

What does

v = ω A^{2} - x^{2}

give if you (wrongly) plug in

x > A

A negative number under the square root, which is unphysical — it signals the mass can never be beyond

x = \pm A

, because that would demand negative KE.

A

is a hard wall set by the energy.

At the two symmetric points

x = + 0.5 A

and

x = - 0.5 A

, how do the energies compare?

They are identical:

U = \frac{1}{2} k (0.5 A)^{2}

and

K = \frac{1}{2} k (A^{2} - 0.25 A^{2})

depend only on

x^{2}

, which is the same for

\pm x

. Energy is symmetric about equilibrium.

Boundary check: is there any point where KE and PE are both at their maximum simultaneously?

No.

K

peaks at

x = 0

where

U = 0

, and

U

peaks at

x = \pm A

where

K = 0

. Their maxima are at opposite locations — they are perfectly out of step.

Recall One-line summary of every trap

The word "energy" hides a sum: $K + U = \frac{1}{2} k A^{2}$ is fixed, but the split flips between the centre (all $K$ ) and the edges (all $U$ ), everything squared scales quadratically, and the $sin^{2} / cos^{2}$ squaring is what makes the energies beat at $2 ω$ .