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CLASS 11 PHYSICS

SHM Equations and Graphs

Displacement, velocity, acceleration, phase, phase difference, x-t graph, v-t graph, a-t graph, ellipse relations, numericals and PYQs.

CBSENEETJEE MainJEE AdvancedIBICSEIGCSEA-Level

1. Displacement Equation in SHM

The displacement equation tells where the particle is at any instant. For sine form, the particle starts from mean position and initially moves toward the positive side.

x = A sin(ωt)

x is displacement from mean position, A is amplitude, ω is angular frequency and t is time. At t = 0, x = 0 and the graph initially rises.

Physical Meaning

The particle does not move uniformly. It is fastest near the mean position and slowest near the extreme. The sine equation captures this smooth repeated motion.

Example

If x = 5 sin(10t) cm, amplitude is 5 cm and angular frequency is 10 rad/s. The body begins at the mean position.

Exam Trap and Common Mistake

Trap: x = A sin(ωt) and x = A cos(ωt) describe the same SHM with different starting points.

Mistake: taking A as total distance between two extremes. The distance between extremes is 2A, not A.

Memory trick: sine form starts from zero; cosine form starts from maximum positive displacement.

2. Velocity Equation Derivation

Velocity is the rate of change of displacement. Differentiating x = A sin(ωt) gives the velocity equation.

Starting with x = A sin(ωt), velocity is dx/dt. Since derivative of sin(ωt) is ω cos(ωt),

v = Aω cos(ωt)

Velocity starts from maximum positive value because cos 0 = 1. This matches the sine-form displacement: at mean position, speed is maximum.

v = Aω sin(ωt + π/2)

This sine-form relation shows that velocity leads displacement by π/2.

Physical Meaning, Example and Mistake

Physical meaning: velocity is maximum at mean position and zero at extremes.

Example: if A = 0.20 m and ω = 5 rad/s, maximum speed is Aω = 1 m/s.

Exam trap: velocity is not always Aω. Aω is only maximum velocity. At general x, use v² = ω²(A² − x²).

3. Acceleration Equation Derivation

Acceleration is the rate of change of velocity. It is always directed towards the mean position.

From v = Aω cos(ωt), differentiating again gives acceleration:

a = −Aω² sin(ωt)

At t = 0, acceleration is zero. Immediately after t = 0, displacement is positive and acceleration becomes negative.

a = −ω²x

Since x = A sin(ωt), acceleration is directly proportional to displacement and opposite in direction.

a = Aω² sin(ωt + π)

This shows acceleration differs in phase from displacement by π.

Physical Meaning, Example and Common Mistake

Physical meaning: when the particle is on the positive side, acceleration pulls it back negative. When it is on the negative side, acceleration is positive.

Example: if ω = 10 rad/s and x = 0.03 m, a = −100 × 0.03 = −3 m/s².

Mistake: saying acceleration is maximum at mean position. Actually acceleration is zero at mean and maximum at extremes.

4. Phase Concept

Phase tells the stage of oscillation. In x = A sin(ωt + φ), the quantity (ωt + φ) is phase.

Definition

Phase decides the current position and direction of motion in a cycle. Two SHMs can have the same amplitude and frequency but different phases.

Physical Meaning

If two students on swings pass the mean position together in the same direction, they are in phase. If one is at the right extreme while the other is at the left extreme, they differ by π.

Exam Trap

Do not compare only positions. Same position can occur with opposite velocities. Full phase comparison needs stage and direction.

5. Phase Difference Between x, v and a

Required Relations

v leads x by π/2 because v = Aω sin(ωt + π/2).
a is opposite in phase to x because a = −ω²x.
a leads x by π or differs by π because a = Aω² sin(ωt + π).
a leads v by π/2 in phase language; equivalently v and a differ by π/2.

Example, Trap and Mistake

Example: at t = 0 for x = A sin(ωt), displacement is zero, velocity is maximum positive and acceleration is zero.

Trap: "lead" depends on the chosen reference equation. Always rewrite as sine functions before comparing phase.

Mistake: saying acceleration and displacement are in same phase because both become maximum at extremes. Their signs are opposite, so they differ by π.

Quantity	Equation	At t = 0	Phase Compared With x
Displacement	x = A sin(ωt)	0, increasing	Reference
Velocity	v = Aω cos(ωt) = Aω sin(ωt + π/2)	Maximum positive	Leads x by π/2
Acceleration	a = −Aω² sin(ωt) = Aω² sin(ωt + π)	0, decreasing	Differs from x by π

6. Graphical Understanding

The graphs below follow the sine-form starting condition: x starts from zero and increases, v starts from maximum positive, and a starts from zero and decreases.

x-t Graph: x = A sin(ωt)

v-t Graph: v = Aω cos(ωt)

a-t Graph: a = −Aω² sin(ωt)

Combined Phase Graph

v-x Ellipse

a-x Straight Line: a = −ω²x

7. Ellipse Form Relations

Eliminating time from SHM equations gives relations between displacement, velocity and acceleration. These are powerful in JEE and graph questions.

x²/A² + v²/(A²ω²) = 1

This is the v-x ellipse. Maximum x is A and maximum speed is Aω.

v² = ω²(A² − x²)

This gives speed at any displacement without time.

a = −ω²x

The a-x graph is a straight line through origin with negative slope.

a²/(A²ω⁴) + v²/(A²ω²) = 1

This is the a-v ellipse. Maximum acceleration is Aω².

Exam trap: these ellipse equations use squared quantities, so they give magnitude relations. Direction or sign must be decided separately from phase or position.

Need Help With SHM Equations?

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8. Important Formula Sheet

x = A sin(ωt)

Sine displacement form; starts from mean position.

v = Aω cos(ωt)

Velocity form; starts from maximum positive.

a = −Aω² sin(ωt)

Acceleration form; opposite to displacement.

v_max = Aω

Maximum speed at mean position.

a_max = Aω²

Maximum acceleration at extremes.

T = 2π/ω

Time period from angular frequency.

9. 40 Solved Numericals

Attempt each problem before opening the solution. These cover equations, phase, graphs, velocity-displacement relations and acceleration relations.

CBSE Easy

1. Identify amplitude

Question: x = 8 sin(4t) cm. Find A and ω.

Show Solution

Given: x = 8 sin(4t) cm.

Formula: Compare with x = A sin(ωt).

Solution: A = 8 cm and ω = 4 rad/s.

Final Answer: A = 8 cm, ω = 4 rad/s.

CBSE Easy

2. Time period

Question: x = 5 sin(10πt) cm. Find T.

Show Solution

Given: ω = 10π rad/s.

Formula: T = 2π/ω.

Solution: T = 2π/(10π) = 0.2 s.

Final Answer: 0.2 s.

NEET Easy

3. Maximum speed

Question: A = 0.1 m and ω = 20 rad/s. Find v_max.

Show Solution

Given: A = 0.1 m, ω = 20 rad/s.

Formula: v_max = Aω.

Solution: v_max = 0.1 × 20 = 2 m/s.

Final Answer: 2 m/s.

NEET Easy

4. Maximum acceleration

Question: A = 0.05 m and ω = 10 rad/s. Find a_max.

Show Solution

Given: A = 0.05 m, ω = 10 rad/s.

Formula: a_max = Aω².

Solution: a_max = 0.05 × 100 = 5 m/s².

Final Answer: 5 m/s².

JEE Main Medium

5. Velocity at time

Question: x = 0.2 sin(5t). Find v at t = 0.

Show Solution

Given: A = 0.2 m, ω = 5 rad/s, t = 0.

Formula: v = Aω cos(ωt).

Solution: v = 0.2 × 5 × cos 0 = 1 m/s.

Final Answer: 1 m/s.

JEE Main Medium

6. Acceleration at time

Question: x = 0.1 sin(10t). Find acceleration at t = π/20 s.

Show Solution

Given: A = 0.1 m, ω = 10 rad/s, ωt = π/2.

Formula: a = −Aω² sin(ωt).

Solution: a = −0.1 × 100 × 1 = −10 m/s².

Final Answer: −10 m/s².

CBSE Medium

7. Speed at displacement

Question: A = 10 cm, x = 6 cm, ω = 5 rad/s. Find speed.

Show Solution

Given: A = 0.10 m, x = 0.06 m, ω = 5 rad/s.

Formula: v = ω√(A² − x²).

Solution: v = 5√(0.01 − 0.0036) = 5 × 0.08 = 0.40 m/s.

Final Answer: 0.40 m/s.

NEET Medium

8. Acceleration from x

Question: If x = 0.04 m and ω = 6 rad/s, find a.

Show Solution

Given: x = 0.04 m, ω = 6 rad/s.

Formula: a = −ω²x.

Solution: a = −36 × 0.04 = −1.44 m/s².

Final Answer: −1.44 m/s².

JEE Main Medium

9. Frequency from equation

Question: x = 2 sin(8πt) cm. Find frequency.

Show Solution

Given: ω = 8π rad/s.

Formula: ω = 2πf.

Solution: f = 8π/(2π) = 4 Hz.

Final Answer: 4 Hz.

JEE Advanced Medium

10. Phase at time

Question: For x = A sin(20t), find phase at t = 0.1 s.

Show Solution

Given: ω = 20 rad/s, t = 0.1 s.

Formula: phase = ωt.

Solution: phase = 20 × 0.1 = 2 rad.

Final Answer: 2 rad.

CBSE Easy

11. Phase difference

Question: What is phase difference between x and v?

Show Solution

Given: x = A sin(ωt), v = Aω sin(ωt + π/2).

Formula: compare sine phases.

Solution: v leads x by π/2.

Final Answer: π/2.

NEET Easy

12. Acceleration phase

Question: What is phase difference between x and a?

Show Solution

Given: a = −ω²x.

Formula: opposite sign means phase difference π.

Solution: a and x are opposite in phase.

Final Answer: π.

JEE Main Medium

13. Speed at half amplitude

Question: Find speed at x = A/2 if v_max = 12 m/s.

Show Solution

Given: x = A/2, v_max = Aω = 12.

Formula: v = v_max√(1 − x²/A²).

Solution: v = 12√(1 − 1/4) = 6√3 m/s.

Final Answer: 6√3 m/s.

JEE Main Medium

14. Acceleration at half amplitude

Question: a_max = 24 m/s². Find |a| at x = A/2.

Show Solution

Given: x = A/2.

Formula: |a| = ω²x.

Solution: |a| = a_max/2 = 12 m/s².

Final Answer: 12 m/s².

JEE Advanced Medium

15. Find amplitude

Question: v_max = 3 m/s and ω = 15 rad/s. Find A.

Show Solution

Given: v_max = 3 m/s, ω = 15 rad/s.

Formula: v_max = Aω.

Solution: A = 3/15 = 0.20 m.

Final Answer: 0.20 m.

JEE Advanced Medium

16. Find angular frequency

Question: a_max = 18 m/s² and A = 0.02 m. Find ω.

Show Solution

Given: a_max = 18, A = 0.02.

Formula: a_max = Aω².

Solution: ω² = 18/0.02 = 900, so ω = 30 rad/s.

Final Answer: 30 rad/s.

NEET Medium

17. Displacement at zero velocity

Question: In SHM, velocity is zero. What is displacement?

Show Solution

Given: v = 0.

Formula: v² = ω²(A² − x²).

Solution: A² − x² = 0, so x = ±A.

Final Answer: Extreme position, x = ±A.

CBSE Medium

18. Acceleration at mean

Question: Find acceleration at x = 0.

Show Solution

Given: x = 0.

Formula: a = −ω²x.

Solution: a = 0.

Final Answer: 0.

JEE Main Medium

19. Time for first maximum x

Question: For x = A sin(ωt), when does x first become A?

Show Solution

Given: sin(ωt) = 1.

Formula: ωt = π/2.

Solution: t = π/(2ω) = T/4.

Final Answer: T/4.

NEET Medium

20. Time for first zero velocity

Question: For x = A sin(ωt), when does velocity first become zero?

Show Solution

Given: v = Aω cos(ωt).

Formula: cos(ωt) = 0.

Solution: first zero at ωt = π/2, so t = T/4.

Final Answer: T/4.

CBSE Medium

21. Equation of acceleration

Question: If x = 4 sin(3t) cm, write acceleration equation in cm/s².

Show Solution

Given: A = 4 cm, ω = 3 rad/s.

Formula: a = −Aω² sin(ωt).

Solution: a = −4 × 9 sin(3t) = −36 sin(3t) cm/s².

Final Answer: a = −36 sin(3t) cm/s².

JEE Main Medium

22. Equation of velocity

Question: If x = 0.05 sin(40t), write v.

Show Solution

Given: A = 0.05 m, ω = 40 rad/s.

Formula: v = Aω cos(ωt).

Solution: v = 0.05 × 40 cos(40t) = 2 cos(40t).

Final Answer: v = 2 cos(40t) m/s.

JEE Advanced Difficult

23. Use v-x relation

Question: A = 0.5 m, ω = 4 rad/s. At x = 0.3 m, find v.

Show Solution

Given: A = 0.5, x = 0.3, ω = 4.

Formula: v = ω√(A² − x²).

Solution: v = 4√(0.25 − 0.09) = 4 × 0.4 = 1.6 m/s.

Final Answer: 1.6 m/s.

JEE Advanced Difficult

24. Find x from v

Question: A = 0.2 m, ω = 10 rad/s, speed = 1 m/s. Find |x|.

Show Solution

Given: A = 0.2, ω = 10, v = 1.

Formula: v² = ω²(A² − x²).

Solution: 1 = 100(0.04 − x²), so x² = 0.03.

Final Answer: |x| = √0.03 m = 0.173 m approximately.

NEET Difficult

25. Ratio of speed

Question: Find v/v_max at x = 3A/5.

Show Solution

Given: x/A = 3/5.

Formula: v/v_max = √(1 − x²/A²).

Solution: v/v_max = √(1 − 9/25) = 4/5.

Final Answer: 4/5.

JEE Main Difficult

26. Ratio of acceleration

Question: Find |a|/a_max at x = 3A/5.

Show Solution

Given: x/A = 3/5.

Formula: |a|/a_max = |x|/A.

Solution: ratio = 3/5.

Final Answer: 3/5.

CBSE Medium

27. Sine phase form

Question: Rewrite v = Aω cos(ωt) in sine form.

Show Solution

Given: cos θ = sin(θ + π/2).

Formula: v = Aω sin(ωt + π/2).

Solution: replace θ by ωt.

Final Answer: v = Aω sin(ωt + π/2).

NEET Medium

28. Acceleration sine form

Question: Rewrite a = −Aω² sin(ωt) as a shifted sine.

Show Solution

Given: sin(θ + π) = −sin θ.

Formula: a = Aω² sin(ωt + π).

Solution: use phase shift π.

Final Answer: a = Aω² sin(ωt + π).

JEE Main Difficult

29. Find A from v and x

Question: At x = 0.06 m, v = 0.8 m/s, and ω = 10 rad/s. Find A.

Show Solution

Given: x = 0.06, v = 0.8, ω = 10.

Formula: A² = x² + v²/ω².

Solution: A² = 0.0036 + 0.64/100 = 0.0100.

Final Answer: A = 0.10 m.

JEE Advanced Difficult

30. Find ω from a and x

Question: At x = 0.05 m, acceleration is −20 m/s². Find ω.

Show Solution

Given: a = −20, x = 0.05.

Formula: a = −ω²x.

Solution: 20 = ω² × 0.05, so ω² = 400.

Final Answer: ω = 20 rad/s.

CBSE Easy

31. Initial values

Question: For x = A sin(ωt), find x, v and a at t = 0.

Show Solution

Given: t = 0.

Formula: x = A sin0, v = Aω cos0, a = −Aω² sin0.

Solution: x = 0, v = Aω, a = 0.

Final Answer: x = 0, v = Aω, a = 0.

NEET Medium

32. Quarter period values

Question: For x = A sin(ωt), find x, v and a at t = T/4.

Show Solution

Given: t = T/4 means ωt = π/2.

Formula: use sine and cosine values.

Solution: x = A, v = 0, a = −Aω².

Final Answer: x = A, v = 0, a = −Aω².

JEE Main Medium

33. Half period values

Question: For x = A sin(ωt), find x, v and a at t = T/2.

Show Solution

Given: ωt = π.

Formula: sinπ = 0, cosπ = −1.

Solution: x = 0, v = −Aω, a = 0.

Final Answer: x = 0, v = −Aω, a = 0.

JEE Advanced Medium

34. Three-quarter period

Question: For x = A sin(ωt), find x, v and a at t = 3T/4.

Show Solution

Given: ωt = 3π/2.

Formula: sin(3π/2) = −1, cos(3π/2) = 0.

Solution: x = −A, v = 0, a = +Aω².

Final Answer: x = −A, v = 0, a = Aω².

NEET Medium

35. Determine graph start

Question: A graph starts from x = 0 and rises. Which form is suitable?

Show Solution

Given: starts from zero and increases.

Formula: x = A sin(ωt).

Solution: sine starts at zero with positive slope.

Final Answer: x = A sin(ωt).

JEE Main Medium

36. Determine cosine start

Question: A body starts from positive extreme. Which displacement form is best?

Show Solution

Given: x = A at t = 0.

Formula: cos0 = 1.

Solution: x = A cos(ωt).

Final Answer: x = A cos(ωt).

CBSE Medium

37. a-x graph slope

Question: If a-x graph has slope −25 s⁻², find ω.

Show Solution

Given: slope = −ω² = −25.

Formula: ω² = 25.

Solution: ω = 5 rad/s.

Final Answer: 5 rad/s.

JEE Advanced Difficult

38. v-x ellipse intercepts

Question: A v-x ellipse has x-intercept 0.2 m and v-intercept 4 m/s. Find ω.

Show Solution

Given: A = 0.2 m, Aω = 4 m/s.

Formula: ω = (Aω)/A.

Solution: ω = 4/0.2 = 20 rad/s.

Final Answer: 20 rad/s.

JEE Advanced Difficult

39. a-v ellipse intercepts

Question: In a-v ellipse, v-intercept is 3 m/s and a-intercept is 12 m/s². Find ω.

Show Solution

Given: Aω = 3, Aω² = 12.

Formula: ω = (Aω²)/(Aω).

Solution: ω = 12/3 = 4 rad/s.

Final Answer: 4 rad/s.

JEE Advanced Difficult

40. Combined values

Question: At a point, x = 0.08 m, v = 0.6 m/s and A = 0.10 m. Find ω.

Show Solution

Given: x = 0.08, v = 0.6, A = 0.10.

Formula: v² = ω²(A² − x²).

Solution: 0.36 = ω²(0.01 − 0.0064) = 0.0036ω². Thus ω² = 100.

Final Answer: ω = 10 rad/s.

10. 50 PYQs and Exam Questions

Answers are hidden to support active recall and exam practice.

CBSE

1. Write displacement equation of SHM.

Use sine form and define symbols.

Show Answer

x = A sin(ωt), where A is amplitude, ω is angular frequency and t is time.

CBSE

2. Derive velocity from x = A sin(ωt).

State final expression.

Show Answer

Differentiate displacement: v = dx/dt = Aω cos(ωt).

CBSE

3. Derive acceleration from velocity.

Use v = Aω cos(ωt).

Show Answer

a = dv/dt = −Aω² sin(ωt) = −ω²x.

CBSE

4. Where is velocity maximum?

State position.

Show Answer

Velocity is maximum at mean position, where x = 0.

CBSE

5. Where is acceleration maximum?

State position.

Show Answer

Acceleration magnitude is maximum at extreme positions, x = ±A.

NEET

6. If x = A sin(ωt), what is v at t = 0?

Choose from zero, Aω, −Aω.

Show Answer

v = Aω cos0 = Aω.

NEET

7. If x = A sin(ωt), what is a at t = 0?

State value.

Show Answer

a = −Aω² sin0 = 0.

NEET

8. Which quantity leads displacement by π/2?

In sine-form SHM.

Show Answer

Velocity leads displacement by π/2.

NEET

9. Phase difference between x and a?

Give radians.

Show Answer

π radians.

NEET

10. v² relation in SHM.

Write the formula.

Show Answer

v² = ω²(A² − x²).

JEE Main

11. x = 4 sin(2πt). Find frequency.

Use ω = 2πf.

Show Answer

ω = 2π, so f = 1 Hz.

JEE Main

12. a-x graph slope is −16. Find ω.

Use slope = −ω².

Show Answer

ω² = 16, so ω = 4 rad/s.

JEE Main

13. At x = A/2, find v/v_max.

Use velocity relation.

Show Answer

v/v_max = √(1 − 1/4) = √3/2.

JEE Main

14. At x = A/2, find |a|/a_max.

Use acceleration relation.

Show Answer

|a|/a_max = 1/2.

JEE Main

15. What graph is v-x?

Name the shape.

Show Answer

It is an ellipse: x²/A² + v²/(A²ω²) = 1.

JEE Advanced

16. What graph is a-x?

State slope.

Show Answer

It is a straight line through origin with slope −ω².

JEE Advanced

17. What graph is a-v?

Write relation.

Show Answer

It is an ellipse: a²/(A²ω⁴) + v²/(A²ω²) = 1.

JEE Advanced

18. When velocity is zero, acceleration is?

Use extremes.

Show Answer

At x = ±A, velocity is zero and acceleration magnitude is maximum Aω², directed toward mean position.

JEE Advanced

19. When acceleration is zero, velocity is?

Use mean position.

Show Answer

At x = 0, acceleration is zero and velocity magnitude is maximum Aω.

JEE Advanced

20. If v-x ellipse intercepts are A and Aω, what is ω?

Use intercept ratio.

Show Answer

ω = velocity intercept / displacement intercept.

21. Explain phase in SHM.

One or two sentences.

Show Answer

Phase describes the stage of oscillation and determines position and direction at a given instant.

22. Why is acceleration opposite to displacement?

Use restoring nature.

Show Answer

The acceleration is caused by restoring force, which always acts toward equilibrium, opposite to displacement.

ICSE

23. What is amplitude?

For graph reading.

Show Answer

Maximum displacement from the mean position.

ICSE

24. How to find period from x-t graph?

State graph method.

Show Answer

Measure time between two consecutive identical phase points, such as crest to next crest.

IGCSE

25. Which graph starts from maximum positive for sine displacement?

x, v or a?

Show Answer

The v-t graph starts from maximum positive.

IGCSE

26. Which graph starts from zero and decreases?

For x = A sin(ωt).

Show Answer

The a-t graph starts from zero and decreases.

A-Level

27. State SHM defining equation.

Use acceleration.

Show Answer

a = −ω²x.

A-Level

28. What is the significance of negative sign?

In a = −ω²x.

Show Answer

It shows acceleration is opposite to displacement and directed toward equilibrium.

Assertion-Reason

29. Assertion: v leads x by π/2. Reason: v = Aω sin(ωt + π/2).

Evaluate.

Show Answer

Both are true and the reason correctly explains the assertion.

Assertion-Reason

30. Assertion: a and x are in same phase. Reason: a = −ω²x.

Evaluate.

Show Answer

Assertion is false. Reason is true and shows they are opposite in phase.

True/False

31. v is maximum at x = 0.

True or false?

Show Answer

True.

True/False

32. a is maximum at x = 0.

True or false?

Show Answer

False. Acceleration is zero at x = 0.

True/False

33. x-t and a-t graphs are opposite in phase.

True or false?

Show Answer

True.

True/False

34. v-x graph is a straight line.

True or false?

Show Answer

False. It is an ellipse.

Conceptual

35. Why does v-t graph begin at maximum?

For x = A sin(ωt).

Show Answer

At t = 0 the particle is at mean position, where speed is maximum.

Conceptual

36. Why does a-t graph initially go negative?

For x = A sin(ωt).

Show Answer

Just after t = 0, x becomes positive, so a = −ω²x becomes negative.

Conceptual

37. Can velocity and acceleration be zero together in SHM?

Ideal nonzero amplitude SHM.

Show Answer

No. At mean acceleration is zero but velocity maximum; at extremes velocity is zero but acceleration nonzero.

Conceptual

38. Can displacement and acceleration be zero together?

When?

Show Answer

Yes, at mean position x = 0 and a = −ω²x = 0.

Difficult Numerical

39. v_max = 5 and a_max = 20. Find A.

Use maximum relations.

Show Answer

A = v_max²/a_max = 25/20 = 1.25 m.

Difficult Numerical

40. v_max = 5 and a_max = 20. Find ω.

Use ratio.

Show Answer

ω = a_max/v_max = 20/5 = 4 rad/s.

Difficult Numerical

41. At x = A/√2, find speed.

In terms of v_max.

Show Answer

v/v_max = √(1 − 1/2) = 1/√2.

Difficult Numerical

42. At x = A/√2, find acceleration magnitude.

In terms of a_max.

Show Answer

|a|/a_max = 1/√2.

Case Study

43. A sensor records x = 0 at t = 0 with positive slope.

Choose sine or cosine model.

Show Answer

x = A sin(ωt) is appropriate because it starts from zero and rises.

Case Study

44. A graph of a against x is a straight line through origin with negative slope.

What does it indicate?

Show Answer

It indicates SHM because a = −ω²x.

Case Study

45. A v-x graph has larger vertical intercept after changing system.

What may have increased?

Show Answer

The maximum speed Aω increased, due to greater amplitude, angular frequency, or both.

Case Study

46. A student says acceleration is highest where velocity is highest.

Correct the statement.

Show Answer

Incorrect. Velocity is highest at mean position, where acceleration is zero. Acceleration is highest at extremes.

Mixed

47. What is v when x = A?

Use v-x relation.

Show Answer

v = 0.

Mixed

48. What is a when x = −A?

Use a = −ω²x.

Show Answer

a = +Aω².

Mixed

49. What is displacement when acceleration is maximum positive?

Use sign.

Show Answer

x = −A because a = −ω²x.

Mixed

50. What is displacement when acceleration is maximum negative?

Use sign.

Show Answer

x = +A.

11. Quick Revision Notes

Equation Chain

x = A sin(ωt)
v = Aω cos(ωt)
a = −Aω² sin(ωt)
a = −ω²x

Graph Starts

x starts from zero and rises.
v starts from maximum positive.
a starts from zero and falls negative.
a-x graph has negative slope.

Exam Tips

Use radians for phase.
Compare phases only after converting to sine form.
Use v² relation when time is not given.
Read signs carefully in acceleration questions.

Common Mistakes

Forgetting ω while differentiating.
Calling v = Aω true at all points.
Missing the negative sign in acceleration.
Drawing a-t graph in phase with x-t graph.

Last Day Formulae

v_max = Aω
a_max = Aω²
T = 2π/ω
x²/A² + v²/(A²ω²) = 1

Memory Tricks

Sine starts at zero.
Velocity leads displacement by one quarter cycle.
Acceleration opposes displacement.
Ellipse means time was eliminated.

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SHM Equations and Graphs

1. Displacement Equation in SHM

Physical Meaning

Example

Exam Trap and Common Mistake

2. Velocity Equation Derivation

Physical Meaning, Example and Mistake

3. Acceleration Equation Derivation

Physical Meaning, Example and Common Mistake

4. Phase Concept

Definition

Physical Meaning

Exam Trap

5. Phase Difference Between x, v and a

Required Relations

Example, Trap and Mistake

6. Graphical Understanding

7. Ellipse Form Relations

Need Help With SHM Equations?

8. Important Formula Sheet

9. 40 Solved Numericals

1. Identify amplitude

2. Time period

3. Maximum speed

4. Maximum acceleration

5. Velocity at time

6. Acceleration at time

7. Speed at displacement

8. Acceleration from x

9. Frequency from equation

10. Phase at time

11. Phase difference

12. Acceleration phase

13. Speed at half amplitude

14. Acceleration at half amplitude

15. Find amplitude

16. Find angular frequency

17. Displacement at zero velocity

18. Acceleration at mean

19. Time for first maximum x

20. Time for first zero velocity

21. Equation of acceleration

22. Equation of velocity

23. Use v-x relation

24. Find x from v

25. Ratio of speed

26. Ratio of acceleration

27. Sine phase form

28. Acceleration sine form

29. Find A from v and x

30. Find ω from a and x

31. Initial values

32. Quarter period values

33. Half period values

34. Three-quarter period

35. Determine graph start

36. Determine cosine start

37. a-x graph slope

38. v-x ellipse intercepts

39. a-v ellipse intercepts

40. Combined values

10. 50 PYQs and Exam Questions

1. Write displacement equation of SHM.

2. Derive velocity from x = A sin(ωt).

3. Derive acceleration from velocity.

4. Where is velocity maximum?

5. Where is acceleration maximum?

6. If x = A sin(ωt), what is v at t = 0?

7. If x = A sin(ωt), what is a at t = 0?

8. Which quantity leads displacement by π/2?

9. Phase difference between x and a?

10. v2 relation in SHM.

11. x = 4 sin(2πt). Find frequency.

12. a-x graph slope is −16. Find ω.

13. At x = A/2, find v/vmax.

14. At x = A/2, find |a|/amax.

15. What graph is v-x?

16. What graph is a-x?

17. What graph is a-v?

18. When velocity is zero, acceleration is?

10. v² relation in SHM.

13. At x = A/2, find v/v_max.

14. At x = A/2, find |a|/a_max.

30. Assertion: a and x are in same phase. Reason: a = −ω²x.

39. v_max = 5 and a_max = 20. Find A.

40. v_max = 5 and a_max = 20. Find ω.