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 (a) 0.01 s (b) 10 s (c) 0.1 s (d) 100 s

Answer: (a). vmax = (2π A)/(T) ⇒ T = 2π Avmax T = 2× 3.14× 7×10⁻³4.4 ≈ (0.044)/(4.4) T = 0.01 s So the answer is (a) 0.01 s.

The maximum velocity of a particle, executing simple harmonic motion… | JEE Main 2006 (AIEEE)

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Velocity & Acceleration in SHM

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7\text{ mm},$ is $4.4\text{ m/s}.$ The period of oscillation is

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Answer: (a)

$v_{max} = \frac{2\pi A}{T} \Rightarrow T = \frac{2\pi A}{v_{max}}$
$T = \frac{2\times 3.14\times 7\times10^{-3}}{4.4} \approx \frac{0.044}{4.4}$
$T = 0.01$ s
So the answer is (a) 0.01 s.

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