Energy in Simple Harmonic Motion | Kumar Physics Classes
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CLASS 11 PHYSICS

Energy in Simple Harmonic Motion

Class 11 Physics notes covering kinetic energy, potential energy, total energy, energy conservation, energy graphs, spring-mass systems, numericals and PYQs.

CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level

Introduction to Energy in SHM

Energy in SHM explains why the particle speeds up near the mean position, slows down near extremes, and still keeps repeating the motion in an ideal system.

Definition and Physical Meaning

In ideal simple harmonic motion, mechanical energy continuously changes form between kinetic energy and potential energy. Kinetic energy belongs to motion; potential energy belongs to displacement from equilibrium. At the mean position, the particle is fastest and kinetic energy is maximum. At extreme positions, speed is zero and potential energy is maximum.

Real-life example: in a spring-block system, when the block passes the natural length position, the spring is neither stretched nor compressed, so spring potential energy is minimum and the block has maximum speed.

Exam Perspective

NEET often asks direct energy ratios such as K:U at x = A/2 or the position where K = U. JEE Main asks graph interpretation and velocity from energy. JEE Advanced can combine springs, equivalent spring constant, energy conservation and phase-based reasoning.

Memory trick: center means speed; extremes mean stored energy. Total energy is the fixed budget.

Kinetic Energy in SHM

Kinetic energy is the energy due to motion. In SHM it is maximum at the mean position and zero at extremes.

Mathematical Derivation

Start from the ordinary kinetic energy formula:

K = ½mv2

Here m is mass and v is instantaneous speed.

For SHM, v2 = ω2(A2 − x2). Substituting this in K = ½mv2 gives:

K = ½mω2(A2 − x2)

Kinetic energy depends on displacement. It is largest when x = 0 and becomes zero when x = ±A.

Maximum, Minimum and Time Period

Maximum K: at mean position x = 0, Kmax = ½mω2A2.

Minimum K: at extremes x = ±A, speed is zero, so K = 0.

If x = A sin(ωt), then v = Aω cos(ωt), so K = ½mA2ω2 cos2(ωt). If x = A cos(ωt), then K is proportional to sin2(ωt). In both choices, the squared trigonometric function repeats after π, so kinetic energy repeats after half the displacement period.

Time Period of K = T/2

Trap: the particle takes time T to repeat its full motion, but kinetic energy repeats twice in one oscillation because speed magnitude is high at every mean crossing.

Potential Energy in SHM

Potential energy is stored energy due to displacement from equilibrium. In a spring-mass SHM, it is spring potential energy.

Derivation

For a spring obeying Hooke's law, F = kx in magnitude. Work stored in stretching from 0 to x is the area under the F-x graph:

U = ½kx2

Since k = mω2, the same formula becomes:

U = ½mω2x2

Potential energy is zero at mean position and maximum at extremes.

Variation With Time

If x = A sin(ωt), then U = ½mω2A2 sin2(ωt). Therefore U ∝ sin2(ωt). If cosine form is chosen, U ∝ cos2(ωt).

Time Period of U = T/2

Graphical meaning: potential energy is maximum at both +A and −A, so the same energy state appears twice in one full oscillation.

Common mistake: thinking PE is negative on the negative side. Since U contains x2, it is positive for both +x and −x.

Total Energy in SHM

Total mechanical energy is the sum of kinetic and potential energy. In ideal SHM, it remains constant.

Derivation

E = K + U

Use K = ½mω2(A2 − x2) and U = ½mω2x2.

Adding them:

E = ½mω2A2

The x2 terms cancel. Total energy depends on mass, angular frequency and amplitude, not on instantaneous displacement or time.

Time Period of Total Energy

Total energy does not oscillate in an ideal system. Its graph with time is a constant horizontal line.

Time Period of E = ∞

Some teachers say total energy has no period because it never changes; in exam language, write "constant, no oscillation" or "infinite period".

Trap: total energy is not maximum at extremes and minimum at mean. It is the same everywhere.

Energy Conservation and Transformation

Energy conservation in SHM means K + U = constant. The form of energy changes, but the total stays fixed.

Mean Position

At x = 0, U = 0 and K is maximum. The particle has maximum speed. In a spring-block system, the spring is at natural length.

Extreme Position

At x = ±A, v = 0 and K = 0. Potential energy is maximum because the spring is maximally stretched or compressed.

Between Mean and Extreme

Both K and U are present. As the block moves away from mean position, K decreases and U increases. While returning, U converts back into K.

Conceptual trap: energy conservation does not mean K and U are separately constant. It means their sum is constant.

Special JEE Concept: When K = U

The position where kinetic energy equals potential energy is one of the most repeated SHM energy concepts.

At displacement x, U/E = x2/A2. If K = U, then each is half of total energy.

U = E/2

So x2/A2 = 1/2.

x = ±A/√2

The two positions are symmetric about the mean position.

Graphical Interpretation

On the energy-displacement graph, K and U curves intersect where each is half of total energy. Those intersections occur at x = +A/√2 and x = −A/√2.

Exam trap: students often write x = A/2. That is wrong. Equal energy occurs at 0.707A approximately, not 0.5A.

Memory trick: equal sharing of energy means square sharing, so the distance involves √2.

Energy Graphs in SHM

Energy graphs are parabolic with displacement and double-frequency with time. Total energy remains a constant line.

Kinetic Energy vs Displacement

xKK = E at mean−A+A

Potential Energy vs Displacement

xUU = 0 at mean−A+A

Total Energy vs Displacement

xEconstant

Kinetic Energy vs Time: Period T/2

tKK = E at mean; repeats every T/2

Potential Energy vs Time: Period T/2

tUU = 0 at mean; repeats every T/2

Total Energy vs Time: No Oscillation

tET = ∞ / constant

Combined Energy Graph vs Displacement

xEnergyEK reaches E at meanU reaches E at extremesU = 0 at mean

Energy Exchange Graph

KUEK + U = E, both stay non-negative

Spring-Mass System Energy

A spring-mass system is the standard energy model of SHM because Hooke's law gives U = ½kx² directly.

Mean Position and Extremes

mean: K max−A: U max+A: U max

Energy Transformation

At the right extreme, the spring is stretched and U is maximum. The block starts moving back, U decreases and K increases. At mean position, U is minimum and K is maximum. The block continues due to inertia and compresses the spring on the other side, converting K back into U.

Exam trap: friction is ignored in ideal SHM. If damping or friction is mentioned, mechanical energy is no longer constant.

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Premium Formula Sheet

K = ½mv2

Kinetic energy from speed.

K = ½mω2(A2 − x2)

Kinetic energy at displacement x.

U = ½kx2

Spring potential energy.

U = ½mω2x2

Potential energy using k = mω².

E = ½mω2A2

Total energy, independent of x and t.

x = ±A/√2 when K = U

Equal energy positions.

Time Period of K = T/2

K varies as sin² or cos².

Time Period of U = T/2

U also varies as squared sine or cosine.

Time Period of E = ∞

Total energy is constant, so there is no oscillation.

40 Solved Numericals

Solutions are hidden. First identify whether the question asks for K, U, total energy, ratio, graph period, or equal-energy position.

CBSE Easy

1. Kinetic energy from speed

Question: A 2 kg particle has speed 3 m/s. Find K.

Show Solution

Given: m = 2 kg, v = 3 m/s.

Formula: K = ½mv2.

Solution: K = ½ × 2 × 9 = 9 J.

Final Answer: 9 J.

CBSE Easy

2. Spring potential energy

Question: k = 100 N/m and x = 0.10 m. Find U.

Show Solution

Given: k = 100 N/m, x = 0.10 m.

Formula: U = ½kx2.

Solution: U = ½ × 100 × 0.01 = 0.5 J.

Final Answer: 0.5 J.

NEET Easy

3. Total energy

Question: m = 1 kg, ω = 10 rad/s, A = 0.20 m. Find E.

Show Solution

Given: m = 1, ω = 10, A = 0.20.

Formula: E = ½mω2A2.

Solution: E = ½ × 1 × 100 × 0.04 = 2 J.

Final Answer: 2 J.

NEET Easy

4. K at mean position

Question: Total energy is 5 J. Find K at mean position.

Show Solution

Given: E = 5 J, x = 0.

Formula: At mean, U = 0 and K = E.

Solution: K = 5 J.

Final Answer: 5 J.

CBSE Easy

5. U at mean position

Question: Find U at x = 0.

Show Solution

Given: x = 0.

Formula: U = ½kx2.

Solution: U = 0.

Final Answer: 0.

NEET Easy

6. K at extreme

Question: At x = A, find K.

Show Solution

Given: x = A.

Formula: K = ½mω2(A2 − x2).

Solution: K = 0.

Final Answer: 0.

JEE Main Medium

7. K at displacement

Question: m = 2 kg, ω = 5 rad/s, A = 0.20 m, x = 0.12 m. Find K.

Show Solution

Given: m = 2, ω = 5, A = 0.20, x = 0.12.

Formula: K = ½mω2(A2 − x2).

Solution: K = 1 × 25 × (0.04 − 0.0144) = 0.64 J.

Final Answer: 0.64 J.

JEE Main Medium

8. U using m and ω

Question: m = 0.5 kg, ω = 8 rad/s, x = 0.10 m. Find U.

Show Solution

Given: m = 0.5, ω = 8, x = 0.10.

Formula: U = ½mω2x2.

Solution: U = ½ × 0.5 × 64 × 0.01 = 0.16 J.

Final Answer: 0.16 J.

NEET Medium

9. K:U at x = A/2

Question: Find K:U at x = A/2.

Show Solution

Given: x/A = 1/2.

Formula: U/E = x2/A2.

Solution: U/E = 1/4 and K/E = 3/4.

Final Answer: K:U = 3:1.

NEET Medium

10. K = U position

Question: Find x when K = U.

Show Solution

Given: K = U.

Formula: U/E = x2/A2 = 1/2.

Solution: x = ±A/√2.

Final Answer: ±A/√2.

JEE Main Medium

11. Amplitude from energy

Question: k = 200 N/m and E = 4 J. Find A.

Show Solution

Given: k = 200, E = 4.

Formula: E = ½kA2.

Solution: 4 = 100A2, so A = 0.20 m.

Final Answer: 0.20 m.

JEE Main Medium

12. Spring constant from energy

Question: E = 1 J, A = 0.10 m. Find k.

Show Solution

Given: E = 1, A = 0.10.

Formula: E = ½kA2.

Solution: 1 = 0.005k, so k = 200 N/m.

Final Answer: 200 N/m.

CBSE Medium

13. Period of kinetic energy

Question: SHM time period is 2 s. Find time period of K.

Show Solution

Given: T = 2 s.

Formula: TK = T/2.

Solution: TK = 1 s.

Final Answer: 1 s.

CBSE Medium

14. Period of potential energy

Question: SHM time period is 0.8 s. Find time period of U.

Show Solution

Given: T = 0.8 s.

Formula: TU = T/2.

Solution: TU = 0.4 s.

Final Answer: 0.4 s.

NEET Medium

15. Period of total energy

Question: What is the time period of total energy?

Show Solution

Given: Ideal SHM.

Formula: E is constant.

Solution: It does not oscillate.

Final Answer: Infinite / no oscillation.

JEE Advanced Medium

16. K fraction

Question: Find K/E at x = 3A/5.

Show Solution

Given: x/A = 3/5.

Formula: K/E = 1 − x2/A2.

Solution: K/E = 1 − 9/25 = 16/25.

Final Answer: 16/25.

JEE Advanced Medium

17. U fraction

Question: Find U/E at x = 3A/5.

Show Solution

Given: x/A = 3/5.

Formula: U/E = x2/A2.

Solution: U/E = 9/25.

Final Answer: 9/25.

NEET Medium

18. K:U at 3A/5

Question: Find K:U at x = 3A/5.

Show Solution

Given: K/E = 16/25, U/E = 9/25.

Formula: Ratio K:U.

Solution: K:U = 16:9.

Final Answer: 16:9.

JEE Main Difficult

19. Speed from energy

Question: E = 10 J, U = 6 J, m = 2 kg. Find speed.

Show Solution

Given: E = 10, U = 6, m = 2.

Formula: K = E − U and K = ½mv².

Solution: K = 4 J. 4 = ½ × 2 × v², so v = 2 m/s.

Final Answer: 2 m/s.

JEE Main Difficult

20. Displacement from U

Question: k = 50 N/m and U = 1 J. Find |x|.

Show Solution

Given: k = 50, U = 1.

Formula: U = ½kx².

Solution: 1 = 25x², so x = 0.20 m.

Final Answer: 0.20 m.

JEE Advanced Difficult

21. Energy doubled

Question: If amplitude doubles, what happens to total energy?

Show Solution

Given: A' = 2A.

Formula: E ∝ A².

Solution: E' = 4E.

Final Answer: Four times.

JEE Advanced Difficult

22. Energy and frequency

Question: If ω doubles for same m and A, what happens to E?

Show Solution

Given: ω' = 2ω.

Formula: E ∝ ω².

Solution: E' = 4E.

Final Answer: Four times.

CBSE Medium

23. K maximum

Question: k = 80 N/m, A = 0.5 m. Find maximum K.

Show Solution

Given: k = 80, A = 0.5.

Formula: Kmax = E = ½kA².

Solution: Kmax = 40 × 0.25 = 10 J.

Final Answer: 10 J.

NEET Medium

24. U maximum

Question: m = 1 kg, ω = 4 rad/s, A = 0.25 m. Find Umax.

Show Solution

Given: m = 1, ω = 4, A = 0.25.

Formula: Umax = ½mω²A².

Solution: Umax = ½ × 16 × 0.0625 = 0.5 J.

Final Answer: 0.5 J.

JEE Main Medium

25. U at A/√2

Question: Total energy is 12 J. Find U at x = A/√2.

Show Solution

Given: x = A/√2.

Formula: U = E/2.

Solution: U = 6 J.

Final Answer: 6 J.

JEE Main Medium

26. K at A/√2

Question: Total energy is 12 J. Find K at x = A/√2.

Show Solution

Given: x = A/√2.

Formula: K = E/2.

Solution: K = 6 J.

Final Answer: 6 J.

NEET Difficult

27. Position where K = 3U

Question: Find |x| if K = 3U.

Show Solution

Given: K = 3U, so E = 4U.

Formula: U/E = x²/A².

Solution: U/E = 1/4, so |x| = A/2.

Final Answer: A/2.

JEE Advanced Difficult

28. Position where U = 3K

Question: Find |x| if U = 3K.

Show Solution

Given: U = 3K, so E = 4K and U/E = 3/4.

Formula: U/E = x²/A².

Solution: x²/A² = 3/4, so |x| = √3A/2.

Final Answer: √3A/2.

JEE Main Difficult

29. Energy from velocity maximum

Question: m = 2 kg and vmax = 5 m/s. Find total energy.

Show Solution

Given: m = 2, vmax = 5.

Formula: E = Kmax = ½mvmax².

Solution: E = ½ × 2 × 25 = 25 J.

Final Answer: 25 J.

JEE Advanced Difficult

30. Energy from acceleration maximum

Question: m = 1 kg, A = 0.2 m, amax = 8 m/s². Find E.

Show Solution

Given: amax = Aω², so ω² = 8/0.2 = 40.

Formula: E = ½mω²A².

Solution: E = ½ × 1 × 40 × 0.04 = 0.8 J.

Final Answer: 0.8 J.

CBSE Medium

31. Number of K cycles

Question: SHM completes 5 oscillations. How many cycles of K occur?

Show Solution

Given: K period = T/2.

Formula: K repeats twice per SHM oscillation.

Solution: 5 oscillations give 10 K cycles.

Final Answer: 10.

NEET Medium

32. Number of U maxima

Question: In one full SHM oscillation, how many times is U maximum?

Show Solution

Given: U maximum at x = +A and x = −A.

Formula: two extremes per cycle.

Solution: U is maximum twice.

Final Answer: 2 times.

JEE Main Medium

33. Number of K maxima

Question: In one oscillation, how many times is K maximum?

Show Solution

Given: K maximum at mean crossing.

Formula: mean is crossed twice per cycle.

Solution: K maximum occurs twice.

Final Answer: 2 times.

CBSE Easy

34. Total energy graph

Question: What is the shape of E vs x graph?

Show Solution

Given: E is constant.

Formula: E = ½mω²A².

Solution: It is a horizontal line.

Final Answer: Horizontal straight line.

NEET Easy

35. U graph shape

Question: What is the shape of U vs x graph?

Show Solution

Given: U = ½kx².

Formula: quadratic in x.

Solution: Upward parabola.

Final Answer: Parabola opening upward.

JEE Main Medium

36. K graph shape

Question: What is the shape of K vs x graph?

Show Solution

Given: K = ½mω²(A² − x²).

Formula: negative quadratic in x.

Solution: Downward parabola.

Final Answer: Parabola opening downward.

JEE Advanced Medium

37. K from E and x

Question: E = 20 J and x = A/4. Find K.

Show Solution

Given: x/A = 1/4.

Formula: K/E = 1 − x²/A².

Solution: K = 20(1 − 1/16) = 20 × 15/16 = 18.75 J.

Final Answer: 18.75 J.

JEE Advanced Medium

38. U from E and x

Question: E = 20 J and x = A/4. Find U.

Show Solution

Given: x/A = 1/4.

Formula: U/E = x²/A².

Solution: U = 20 × 1/16 = 1.25 J.

Final Answer: 1.25 J.

JEE Advanced Difficult

39. Speed at equal energy

Question: vmax = 10 m/s. Find speed where K = U.

Show Solution

Given: K = E/2.

Formula: K/Kmax = v²/vmax².

Solution: v = vmax/√2 = 10/√2.

Final Answer: 5√2 m/s.

JEE Advanced Difficult

40. Combined energy

Question: k = 300 N/m, A = 0.2 m, x = 0.1 m. Find K and U.

Show Solution

Given: k = 300, A = 0.2, x = 0.1.

Formula: E = ½kA², U = ½kx², K = E − U.

Solution: E = 150 × 0.04 = 6 J. U = 150 × 0.01 = 1.5 J. K = 4.5 J.

Final Answer: K = 4.5 J, U = 1.5 J.

50 PYQs and Exam Questions

This bank includes school-board, NEET, JEE, international, graph, assertion-reason, true/false and case-study questions. Answers are hidden.

CBSE

1. Define kinetic energy in SHM.

State where it is maximum.

Show Answer
Kinetic energy is ½mv². It is maximum at mean position.
CBSE

2. Define potential energy in spring SHM.

Write the formula.

Show Answer
Potential energy is stored energy due to displacement: U = ½kx².
CBSE

3. Write total energy of SHM.

Use m, ω and A.

Show Answer
E = ½mω²A².
CBSE

4. Why is total energy constant?

Ideal SHM.

Show Answer
Because kinetic and potential energy transform into each other without loss, so K + U remains constant.
CBSE

5. Where is potential energy zero?

For spring SHM.

Show Answer
At mean position, x = 0.
NEET

6. Period of kinetic energy.

If SHM period is T.

Show Answer
Time period of kinetic energy is T/2.
NEET

7. Period of potential energy.

If SHM period is T.

Show Answer
Time period of potential energy is T/2.
NEET

8. Period of total energy.

Ideal SHM.

Show Answer
Total energy is constant, so it has no oscillation or infinite period.
NEET

9. At x = A/2, find K:U.

Energy ratio.

Show Answer
K:U = 3:1.
NEET

10. At what x is K = U?

Give both positions.

Show Answer
x = ±A/√2.
JEE Main

11. Shape of K-x graph.

Identify the curve.

Show Answer
Downward opening parabola.
JEE Main

12. Shape of U-x graph.

Identify the curve.

Show Answer
Upward opening parabola.
JEE Main

13. Shape of E-x graph.

Identify the curve.

Show Answer
Horizontal straight line.
JEE Main

14. If amplitude doubles, energy becomes?

Same system.

Show Answer
Four times because E ∝ A².
JEE Main

15. If k doubles with same A, energy becomes?

Spring SHM.

Show Answer
Energy doubles because E = ½kA².
JEE Advanced

16. Derive K in terms of x.

State final result.

Show Answer
Using v² = ω²(A² − x²), K = ½mω²(A² − x²).
JEE Advanced

17. Derive total energy.

Use K and U.

Show Answer
K + U = ½mω²(A² − x²) + ½mω²x² = ½mω²A².
JEE Advanced

18. If K = 3U, find x.

Magnitude only.

Show Answer
|x| = A/2.
JEE Advanced

19. If U = 3K, find x.

Magnitude only.

Show Answer
|x| = √3A/2.
JEE Advanced

20. How many times per cycle is K maximum?

Explain briefly.

Show Answer
Twice, because the particle crosses mean position twice.
IB

21. Describe energy exchange in SHM.

Use K and U.

Show Answer
K and U continuously transform into each other while total energy remains constant.
IB

22. Why is U never negative for spring SHM?

Use formula.

Show Answer
U = ½kx², and x² is always non-negative.
ICSE

23. Where is speed zero?

Energy view.

Show Answer
At extreme positions, where K = 0 and U is maximum.
ICSE

24. Where is speed maximum?

Energy view.

Show Answer
At mean position, where K is maximum.
IGCSE

25. What happens to energy if friction exists?

Real oscillator.

Show Answer
Mechanical energy decreases with time, usually becoming heat.
IGCSE

26. What is conserved in ideal SHM?

Name the quantity.

Show Answer
Total mechanical energy.
A-Level

27. Explain why K has period T/2.

Use squared sine/cosine.

Show Answer
K is proportional to sin²(ωt) or cos²(ωt), which repeats after π phase, i.e. T/2.
A-Level

28. Explain why U has period T/2.

Use squared displacement.

Show Answer
U ∝ x², so +x and −x give same U. Hence U repeats twice per SHM cycle.
Assertion-Reason

29. Assertion: Total energy in ideal SHM is constant. Reason: K and U transform into each other without loss.

Evaluate.

Show Answer
Both are true, and the reason correctly explains the assertion.
Assertion-Reason

30. Assertion: Kinetic energy period is T. Reason: velocity period is T.

Evaluate.

Show Answer
Assertion is false. Reason is true for velocity, but K depends on v² and has period T/2.
True/False

31. Total energy is maximum at extremes.

True or false?

Show Answer
False. Total energy is constant everywhere.
True/False

32. Potential energy is maximum at both extremes.

True or false?

Show Answer
True.
True/False

33. Kinetic energy can be negative.

True or false?

Show Answer
False. K = ½mv² is never negative.
True/False

34. U at x = −A equals U at x = +A.

True or false?

Show Answer
True, because U depends on x².
Conceptual

35. Why does K become zero at extremes?

Energy reasoning.

Show Answer
At extremes, velocity is zero, so K = ½mv² = 0.
Conceptual

36. Why is U maximum at extremes?

Energy reasoning.

Show Answer
Displacement magnitude is maximum at extremes and U = ½kx².
Conceptual

37. Why is total energy independent of x?

Use cancellation.

Show Answer
In K + U, the x² term in K cancels the x² term in U, leaving ½mω²A².
Conceptual

38. Why does total energy depend on amplitude?

Physical explanation.

Show Answer
Larger amplitude means greater maximum stretch/compression and greater maximum speed, so total energy increases as A².
Difficult Numerical

39. E = 18 J, K = 10 J. Find U.

Use conservation.

Show Answer
U = E − K = 8 J.
Difficult Numerical

40. E = 18 J, U = 2 J, m = 2 kg. Find speed.

Use K = E − U.

Show Answer
K = 16 J. 16 = ½ × 2 × v², so v = 4 m/s.
Difficult Numerical

41. At x = A/3, find U/E.

Ratio question.

Show Answer
U/E = 1/9.
Difficult Numerical

42. At x = A/3, find K/E.

Ratio question.

Show Answer
K/E = 8/9.
Case Study

43. A spring block passes mean position.

Which energy is maximum?

Show Answer
Kinetic energy is maximum.
Case Study

44. A spring block is at maximum compression.

Which energy is maximum?

Show Answer
Potential energy is maximum.
Case Study

45. A graph of total energy with time slopes downward.

What does it indicate?

Show Answer
The oscillator is not ideal; energy is being lost due to damping/friction.
Case Study

46. K and U curves intersect on an energy graph.

What is displacement?

Show Answer
x = ±A/√2.
Graph-Based

47. K-x graph touches zero at which points?

State positions.

Show Answer
At x = +A and x = −A.
Graph-Based

48. U-x graph has minimum where?

State position.

Show Answer
At x = 0.
Graph-Based

49. K-time graph has how many maxima in one SHM period?

Graph interpretation.

Show Answer
Two maxima.
Graph-Based

50. E-time graph is what shape?

Ideal SHM.

Show Answer
A horizontal straight line.

Common Mistakes in Energy in SHM

Total Energy vs Kinetic Energy

Total energy is constant. Kinetic energy changes with position and time. Do not use K = E except at mean position.

Wrong Time Period

K and U have period T/2, not T. They repeat twice because squared quantities do not distinguish positive and negative directions.

K = U Error

The correct position is x = ±A/√2, not A/2.

Graph Mistakes

K-x opens downward, U-x opens upward, and E-x is horizontal. Do not draw all three as sine waves.

Sign Error

Energy is scalar. K and U are not negative in ideal spring SHM.

Damping Trap

If friction, air resistance or damping is present, total mechanical energy decreases. Ideal conservation no longer applies.

Revision Notes

Quick Revision Notes

  • K is maximum at mean.
  • U is maximum at extremes.
  • E = K + U is constant.
  • K and U repeat every T/2.
  • Total energy has no oscillation.

Formula Sheet

  • K = ½mv²
  • K = ½mω²(A² − x²)
  • U = ½kx²
  • U = ½mω²x²
  • E = ½mω²A²

Most Important Concepts

  • K + U = constant.
  • K = U at x = ±A/√2.
  • U/E = x²/A².
  • K/E = 1 − x²/A².
  • Energy is scalar.

Exam Tips

  • For ratios, divide by total energy.
  • For graph questions, identify axis first.
  • Use x/A to avoid unnecessary calculation.
  • Check if the system is ideal.

Last Day Revision Notes

  • Mean: K max, U min.
  • Extreme: K zero, U max.
  • K graph with x is inverted parabola.
  • U graph with x is upward parabola.
  • E graph is horizontal line.

Memory Tricks

  • Center is speed.
  • Ends are stored energy.
  • Total is budget.
  • Squares halve the period.
  • Equal energy means √2.

If Energy in SHM is not clear and you are looking for a Physics Tutor, contact Kumar Sir.

Phone: +91-9958461445   Email: kumarsirphysics@gmail.com   Website: https://kumarphysicsclasses.com

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