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Velocity and energy in SHM: Class 11 practice

In simple harmonic motion x = A sin ωt, speed at displacement x is v = ω√(A² − x²), maximum speed ωA at the mean position, and total energy E = ½mω²A² (shared between KE and PE).. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Simple pendulum and spring

The idea behind velocity and energy in shm

In simple harmonic motion x = A sin ωt, speed at displacement x is v = ω√(A² − x²), maximum speed ωA at the mean position, and total energy E = ½mω²A² (shared between KE and PE).

Step-by-step method

  1. Find ω = 2π/T or √(k/m).
  2. Use v = ω√(A² − x²).
  3. E = ½mω²A²; KE = E − ½mω²x².

Worked example 1 (Basic)

A body in SHM has amplitude 9 cm and ω = 11 rad/s. Find its maximum speed.

  1. v = ωA = 11 × 0.09
  2. Answer: 0.99 m/s

Worked example 2 (Standard)

Amplitude 3 cm, ω = 5 rad/s. Find the speed at x = 1.2 cm.

  1. v = 5√(0.03² − 0.012²)
  2. Answer: 0.137 m/s

Worked example 3 (Challenge)

A 0.5 kg mass oscillates with A = 10 cm, ω = 12 rad/s. Find total energy and KE at x = 7 cm.

  1. E = ½ × 0.5 × 12² × 0.1² = 0.36 J
  2. KE = 0.184 J
  3. Answer: E = 0.36 J, KE = 0.184 J

Common mistakes and misconceptions

  • Using A instead of √(A² − x²)
  • Forgetting ω is in rad/s

Practice questions

  1. (Basic) A body in SHM has amplitude 5 cm and ω = 2 rad/s. Find its maximum speed.
  2. (Basic) A body in SHM has amplitude 3 cm and ω = 9 rad/s. Find its maximum speed.
  3. (Standard) Amplitude 3 cm, ω = 8 rad/s. Find the speed at x = 1.8 cm.
  4. (Standard) Amplitude 4 cm, ω = 8 rad/s. Find the speed at x = 2.4 cm.
  5. (Challenge) A 1 kg mass oscillates with A = 4 cm, ω = 2 rad/s. Find total energy and KE at x = 2 cm.
  6. (Challenge) A 0.5 kg mass oscillates with A = 9 cm, ω = 7 rad/s. Find total energy and KE at x = 4.5 cm.

Answer key with full working

  • Q1: v = ωA = 2 × 0.05 → Answer: 0.1 m/s
  • Q2: v = ωA = 9 × 0.03 → Answer: 0.27 m/s
  • Q3: v = 8√(0.03² − 0.018²) → Answer: 0.192 m/s
  • Q4: v = 8√(0.04² − 0.024²) → Answer: 0.256 m/s
  • Q5: E = ½ × 1 × 2² × 0.04² = 0.003 J → KE = 0.002 J → Answer: E = 0.003 J, KE = 0.002 J
  • Q6: E = ½ × 0.5 × 7² × 0.09² = 0.099 J → KE = 0.074 J → Answer: E = 0.099 J, KE = 0.074 J

Syllabus note

This skill is listed under "Oscillations" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is velocity and energy in shm in?

"Oscillations", Class 11 Physics (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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