SHM — Energy + Spring Systems
SHM
42
JEE Qs
8%
Hard
75
min
Always correctly identify the equilibrium position of the oscillating body first, as all displacements and potential energy calculations in SHM must be referenced from this point.
🧮 Key Formulas
✅ Key Points for JEE
- 1Mechanical energy (sum of kinetic and potential energy) is conserved in ideal SHM due to the conservative nature of the spring force.
- 2Kinetic energy is maximum at the mean position (x=0) and zero at the extreme positions (x=±A). Potential energy is maximum at the extreme positions and zero at the mean position (if PE is referenced to the mean position).
- 3Total mechanical energy of an SHM system depends only on its amplitude (A), spring constant (k), and mass (m) (through ω), not on its instantaneous position.
- 4When gravity acts on a vertical spring-mass system, the mean position (equilibrium position) shifts downwards. SHM occurs about this new equilibrium, and all displacement 'x' in SHM equations (including potential energy calculations) must be measured from this equilibrium position.
- 5Effectively combining springs (series/parallel) is crucial for calculating the equivalent spring constant (k_eq) of complex systems before finding angular frequency or time period.
⚠️ Common Mistakes
- ✕Incorrectly measuring displacement 'x' for potential energy (1/2 kx^2) or other SHM equations. 'x' must always be measured from the *equilibrium position* of SHM, not necessarily the natural length of the spring.
- ✕Confusing the roles of kinetic and potential energy: Forgetting that KE is zero at extremes and PE is max at extremes, or vice-versa.
- ✕Errors in calculating equivalent spring constants for systems with multiple springs, especially when distinguishing between series and parallel arrangements.
- ✕Neglecting the shift in equilibrium position due to gravity in vertical spring-mass systems and using the natural length as the reference for SHM calculations.
📝 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 6: Work, Energy and Power
- Class 11 Physics Ch 14: Oscillations