Energy in SHM — KE + PE = constant
SHM
8
JEE Qs
8%
Hard
60
min
Master the conservation of mechanical energy in SHM and its implications for finding velocity or position at any point, especially maximum/minimum energy instances.
🧮 Key Formulas
✅ Key Points for JEE
- 1Total mechanical energy (E = KE + PE) in SHM remains constant throughout the motion, assuming no damping forces.
- 2Energy continuously transforms between kinetic (KE) and potential (PE); KE is maximum at the equilibrium position (x=0) and zero at extreme positions (x=±A), while PE is zero at equilibrium and maximum at extreme positions.
- 3The total energy of an SHM system is directly proportional to the square of its amplitude (E ∝ A^2) and depends on the system's properties (mass 'm' and angular frequency 'ω' or force constant 'k').
- 4Graphs of KE vs. position (x) and PE vs. position (x) are parabolic. They intersect at x = ±A/√2, where KE = PE = E/2.
- 5The average kinetic energy and average potential energy over one complete cycle of SHM are both equal to half of the total mechanical energy (KE_avg = PE_avg = E/2).
⚠️ Common Mistakes
- ✕Confusing displacement 'x' with amplitude 'A' when calculating instantaneous KE or PE, or when using 'x' instead of 'A' for total energy.
- ✕Incorrectly assuming the general force constant 'k' in PE = 1/2 kx^2 is always the spring constant, without understanding the relation k = mω^2 for any SHM.
- ✕Forgetting that total energy remains constant, especially when considering energy at different points in the oscillation cycle.
- ✕Not using the correct form of potential energy for different SHM systems (e.g., 1/2 kx^2 for spring, mgh for pendulum approximated as 1/2 mg(θ^2)l for small angles).
📝 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 14: Oscillations