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PhysicsOscillations & SHM
The sum of kinetic energy and potential energy of a simple pendulum bob is 0·02 joule. The speed of the simple pendulum bob at equilibrium position is approximately :

(Consider mass of the bob = 20 g)
Options
1
0·2 m/s
2
1·41 m/s
3
14·1 m/s
4
2·0 m/s
Correct Answer
Option 2 : 1·41 m/s
Step-by-Step Solution
1

Key concept — equilibrium position:

At the equilibrium (mean) position of a pendulum, the bob is at its lowest point. Potential energy (measured from this point) = 0.

Therefore: All energy = Kinetic Energy

2

Apply energy conservation:

Total energy = KE + PE = 0.02 J (given)

At equilibrium: PE = 0, so KE = 0.02 J

$$\frac{1}{2}mv^2 = 0.02 \text{ J}$$

3

Solve for v:

Mass m = 20 g = 0.02 kg

$$v^2 = \frac{2 \times 0.02}{0.02} = \frac{0.04}{0.02} = 2$$

$$v = \sqrt{2} \approx \mathbf{1.41 \text{ m/s}}$$ ✓

Theory: Simple Harmonic Motion & Energy
1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a special type of periodic motion where the restoring force on the body is directly proportional to its displacement from the equilibrium position and is always directed towards the equilibrium. Mathematically: F = −kx, where k is the force constant and x is displacement. The negative sign indicates the force is always opposite to displacement.

Displacement in SHM: x = A sin(ωt + φ), where A is amplitude, ω is angular frequency, and φ is initial phase. Velocity: v = Aω cos(ωt + φ) = ω√(A² − x²). Acceleration: a = −ω²x (always towards equilibrium).

2. Energy in SHM

The total mechanical energy in SHM is constant and depends only on amplitude:

\(KE = \dfrac{1}{2}m\omega^2(A^2 - x^2)\)

\(PE = \dfrac{1}{2}m\omega^2 x^2\)

\(E_{total} = KE + PE = \dfrac{1}{2}m\omega^2 A^2\)

📌 At mean position (x = 0): KE = max = E_total, PE = 0

📌 At extreme (x = ±A): KE = 0, PE = max = E_total

📌 At x = A/√2: KE = PE = E_total/2

📌 Total energy ∝ A² (proportional to square of amplitude)

3. Simple Pendulum

A simple pendulum consists of a point mass (bob) suspended by a massless, inextensible string of length L from a fixed point. For small angular displacements (θ < 10°), it undergoes SHM with:

\(T = 2\pi\sqrt{\dfrac{L}{g}}\)

\(\omega = \sqrt{\dfrac{g}{L}}\)

The time period is independent of: (1) mass of the bob, (2) amplitude (for small angles), (3) material of bob. It depends only on: length L and local gravity g. This isochronism (equal time periods) was first observed by Galileo.

4. Energy of Simple Pendulum

For a pendulum bob at angular displacement θ, the height above the lowest point is h = L(1 − cosθ). PE = mgh = mgL(1 − cosθ). For small angles, PE ≈ ½mgLθ² ≈ ½mω²x², which is the SHM form. At equilibrium (θ = 0): PE = 0, KE = maximum = Total E. At extreme (θ = θ_max): KE = 0, PE = Total E = mgL(1 − cosθ_max).

5. Maximum Velocity in SHM

\(v_{max} = A\omega = A\sqrt{\dfrac{g}{L}}\) (for pendulum)

The maximum velocity always occurs at the equilibrium position. In this problem, v_max = √2 ≈ 1.41 m/s. We can find the amplitude if we know ω (i.e., if L is given): A = v_max/ω.

6. Factors Affecting Time Period

📌 Increase in length L → T increases (T ∝ √L)

📌 Increase in g → T decreases (T ∝ 1/√g)

📌 On Moon (g = g/6): T_moon = √6 × T_earth

📌 In a falling lift (effective g = 0): T → ∞ (pendulum stops)

📌 In an accelerating lift (upward): effective g increases → T decreases

Frequently Asked Questions
1. Why is PE zero at equilibrium position?
PE in SHM is measured relative to the equilibrium position. At equilibrium, displacement x = 0, so PE = ½mω²x² = 0. All mechanical energy is kinetic at this point. This is why the bob moves fastest at equilibrium.
2. What is the total energy here and why?
Total energy = KE + PE = 0.02 J (given). Since total energy is constant throughout SHM, and at equilibrium PE = 0, all 0.02 J must be kinetic energy at the equilibrium position.
3. How do we get v = √2 m/s?
½mv² = 0.02 J. m = 20 g = 0.02 kg. v² = 2×0.02/0.02 = 2. v = √2 ≈ 1.414 ≈ 1.41 m/s. The numbers work out neatly because m = 0.02 kg and E = 0.02 J cancel to give v² = 2.
4. Where is the pendulum bob slowest?
The bob is slowest (v = 0) at the extreme positions — the turning points where displacement = amplitude. At these points, all energy is PE and KE = 0. The bob momentarily stops before swinging back.
5. Does the mass of the bob affect the time period?
No. T = 2π√(L/g) is independent of mass. However, mass does affect the energy for a given amplitude (E = ½mω²A²) and the speed for given energy (v = √(2E/m)). Mass matters for energy and force calculations but not for time period.
6. At what displacement is KE = PE?
KE = PE when ½mω²(A²−x²) = ½mω²x², giving A²−x² = x², so x = A/√2. At this displacement, each equals E_total/2 = 0.01 J.
7. How does total energy depend on amplitude?
E = ½mω²A². Total energy is proportional to the square of amplitude. If amplitude doubles, total energy quadruples. If total energy is 0.02 J, we can find amplitude A = √(2E/mω²) if ω is known.
8. What is the velocity at displacement x from equilibrium?
v = ω√(A²−x²). At equilibrium (x=0): v = ωA = v_max. At extreme (x=A): v = 0. Using energy: ½mv² = E − ½mω²x², giving v = √(v²_max − ω²x²).
9. What happens to a pendulum in a free-falling lift?
In a free-falling lift, effective g = 0. T = 2π√(L/0) → ∞. The pendulum does not oscillate at all. This is because the restoring force (component of gravity along arc) also becomes zero in the free-fall frame.
10. Is the pendulum formula T = 2π√(L/g) valid for large amplitudes?
No. It is valid only for small amplitudes (θ < 10°) where sinθ ≈ θ (in radians). For large amplitudes, the period is longer than 2π√(L/g). The exact formula involves an elliptic integral and is not required for NEET.
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