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  4. The maximum velocity of a particle, executing simple harmonic motion with an amplitude 7 mm is 4.4 m s- 1 . The period of oscillation is

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Q. The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 \, mm$ is $4.4 \, m \, s^{- 1}$ . The period of oscillation is

NTA AbhyasNTA Abhyas 2022

Solution:

Maximum velocity $\left(\nu\right)_{\text{m}}=\text{a}\omega =\text{a}\left(\frac{2 \pi }{\text{T}}\right)$
$\therefore $ $\text{T}=\frac{2 \pi \text{a}}{\left(\nu\right)_{\text{m}}}=2\times \frac{22}{7}\times \frac{\left(7 \times \left(\text{10}\right)^{- 3}\right)}{4 \cdot 4}$
$T=10^{- 2}=0.01 \, s$