AimMent
Exam level2 past questions

Relation between Circular Motion and SHM

Understand how SHM can be viewed as the projection of uniform circular motion and how perpendicular SHMs combine.

Why this shows up in the exam

NEET may test your ability to connect circular motion concepts with oscillatory motion.

How NEET tests this

ApplicationNumerical

Learn the idea

A simple harmonic motion is exactly the shadow (projection) of a point moving with uniform speed on a circle; the amplitude is the radius of that circle, found by the Pythagorean sum of the sine and cosine coefficients.

🧠 Memory hook: Think of a spinning wheel casting a shadow on the floor – the shadow’s length (amplitude) is the wheel’s radius found by the Pythagorean ‘shadow‑sum’ √(A²+B²).

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • Uniform circular motion: radius R, angular speed ω, position vector rotates uniformly
  • Projection of the radius vector on any diameter gives x=R cos(ωt+φ) – a SHM
  • Any expression A sin ωt+B cos ωt can be written as R sin(ωt+φ) with R=√(A²+B²)
  • The phase φ is given by tan φ = B/A (or A/B depending on chosen form)
  • Two perpendicular SHMs of equal ω and equal amplitude combine to give uniform circular motion; unequal amplitudes give an ellipse
  • Frequency f = ω/2π and period T = 2π/ω

How to approach it

  1. 1Identify the displacement as A sin ωt + B cos ωt
  2. 2Compute amplitude R = √(A² + B²) – this is the radius of the underlying circle
  3. 3If required, find the phase φ using tan φ = B/A (or A/B) to write a single sinusoid
  4. 4For projection questions, relate ω to the given period (ω = 2π/T) and use the sign of sin/cos to decide initial position and sense of rotation

Worked example — watch it click

The displacement of a particle is given by x = 3 sin (5πt) + 4 cos (5πt) The amplitude of particle is:

  • A)3
  • B)4
  • ✅5
  • D)7

The concept behind this problem

The problem asks for the maximum displacement of a motion written as a sum of sine and cosine terms, which directly tests the conversion to a single sinusoid and the Pythagorean rule for amplitude.

Step by step

  1. 1x = 3sin(5πt) + 4cos(5πt).
  2. 2This is of form x = Asin(ωt) + Bcos(ωt).
  3. 3Amplitude = √(A² + B²) = √(3² + 4²) = √(9 + 16) = √25 = 5.

Watch out

Students often pick 4 (the larger coefficient) as the amplitude instead of the correct √(3²+4²)=5.

Common slip-ups that cost marks

  • •Treating the larger coefficient as the amplitude – the true amplitude is the vector sum √(A²+B²)
  • •Skipping the phase calculation and assuming the motion starts from zero displacement
  • •Assuming any two perpendicular SHMs automatically give a circle; they must have equal amplitudes for a perfect circle

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Practise it

These are real questions from past NEET papers that test this exact idea.

Question 1 of 2NEET 1999

The displacement of a particle is given by x = 3 sin (5πt) + 4 cos (5πt) The amplitude of particle is:

Push further

More challenging

7 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.

Question 1 of 7

A particle undergoes uniform circular motion with a radius of 5 cm. Its period of revolution is 8 seconds. If the particle starts its motion from the positive x-axis and revolves counter-clockwise, what is the equation for the x-projection of its position as a function of time?

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