Angular SHM — Physical pendulum
SHM
8
JEE Qs
8%
Hard
75
min
Always clearly identify the pivot point and the center of mass, then correctly calculate the moment of inertia about the pivot using the Parallel Axis Theorem and the distance 'd' to the COM.
🧮 Key Formulas
✅ Key Points for JEE
- 1A physical pendulum undergoes angular SHM for small angular displacements (θ) from its stable equilibrium position.
- 2The restoring torque (τ) is generated by gravity and acts to bring the pendulum back to its equilibrium. It's crucial to identify the pivot point and the center of mass (COM) correctly.
- 3The moment of inertia (I) must be calculated about the pivot point, not necessarily the COM. Use the Parallel Axis Theorem if the COM is not the pivot.
- 4The distance 'd' in the time period formula is the perpendicular distance from the pivot point to the center of mass of the pendulum.
- 5The small angle approximation (sinθ ≈ θ) is fundamental for the motion to be simple harmonic. Without it, the motion is oscillatory but not SHM.
⚠️ Common Mistakes
- ✕Using the moment of inertia about the center of mass (I_COM) directly in the time period formula instead of the moment of inertia about the pivot point (I_pivot), without applying the Parallel Axis Theorem.
- ✕Incorrectly identifying 'd' as the length of the pendulum or the distance from the pivot to a random point, instead of the distance from the pivot to the center of mass.
- ✕Failing to make the small angle approximation (sinθ ≈ θ), which is necessary for the motion to be classified as SHM.
- ✕Sign errors in the restoring torque calculation, which can lead to an incorrect differential equation or misinterpretation of stable equilibrium.
📝 Practice Questions
See allQ36.A particle oscillates along the x-axis according to the law, x(t) = x0 sin2 ( 2t ) where x0 = 1 m . The kinetic energy (K) of the particle as a function of x is correctly represented by the graph (1) (2) (3) (4)
Q28.Two bodies A and B of equal mass are suspended from two massless springs of spring constant k1 and k2 , respectively. If the bodies oscillate vertically such that their amplitudes are equal, the ratio of the maximum velocity of A to the maximum velocity of B is (1) k1 (2) k2 √k1k2 (3) (4) k2 k1 √k2k1
Q41.A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is yπ × 10−2 s, where the value of y is (Acceleration due to gravity, g = 10 m/s2 , density of water = 103 kg/m3 ) 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) 6 (2) 2 (3) 4 (4) 1
Q36.A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distance and displacement covered by the particle in 12.5 s , then D is d (1) 16 (2) 10 5 (3) 15 (4) 25 4
Q35.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : Knowing initial position x0 and initial momentum p0 is enough to determine the position and momentum at any time t for a simple harmonic motion with a given angular frequency ω. Reason (R): The amplitude and phase can be expressed in terms of x0 and p0 . In the light of the above statements, choose the correct answer from the options given below : (1) (A) is false but (R) is true (2) (A) is true but (R) is false (3) Both (A) and (R) are true but (R) is NOT the (4) Both (A) and (R) are true and (R) is the correct correct explanation of (A) explanation of (A)
Q24.The displacement of a particle executing SHM is given by x = 10 sin (wt + π3 )m. The time period of motion is 3.14 s. The velocity of the particle at t = 0 is ______ m/s.
NCERT Chapters
- Class 11 Physics Ch 7: System of Particles and Rotational Motion
- Class 11 Physics Ch 14: Oscillations