JEE Main · 2006 · 30 Apr 2006 · Offline · Elasticity & SHM

The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm, is

4.4 m / s

. The period of oscillation is

Answer: (B)

Solution

For simple harmonic motion the maximum speed is

v_{m a x} = A ω

. With amplitude

A = 7 mm = 7 \times 1 0^{- 3} m

and

v_{m a x} = 4.4 m/s

, the angular frequency is

ω = \frac{v _{m a x}}{A} = \frac{4.4}{7 \times 1 0 ^{- 3}} \approx 628 rad/s

. The period is then

T = \frac{2 π}{ω} = \frac{2 π}{628} \approx 0.01 s

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option