JEE Main · 2007 · 29 Apr 2007 · Offline · Elasticity & SHM

A particle of mass

m

executes simple harmonic motion with amplitude 'a' and frequency 'v'. The average kinetic energy during its motion from the position of equilibrium to the end is

(A) $π^{2} m a^{2} v^{2}$ ✓ Correct
(B) $\frac{1}{4} π^{2} m a^{2} v^{2}$
(C) $4 π^{2} m a^{2} v^{2}$
(D) $2 π^{2} m a^{2} v^{2}$

Answer: (A)

Solution

Writing the frequency as

v

, the angular frequency is

ω = 2 π v

and the total energy of the motion is

E = \frac{1}{2} m ω^{2} a^{2} = 2 π^{2} m a^{2} v^{2}

. As the particle travels from the equilibrium position out to the extreme, its kinetic energy falls steadily from the full value

E

to zero. Averaged over this stretch the kinetic energy is

\frac{E}{2} = π^{2} m a^{2} v^{2}

JEE Main · 2007 · 29 Apr 2007 · Offline · Single correct option

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