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(A) For a particle executing simple harmonic motion $(SHM)$, the maximum velocity is given by $v_{max} = A\omega$ and the maximum acceleration is give

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$\pi \,s$

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(A) For a particle executing simple harmonic motion $(SHM)$, the maximum velocity is given by $v_{max} = A\omega$ and the maximum acceleration is given by $a_{max} = A\omega^2$, where $A$ is the amplitude and $\omega$ is the angular frequency.Given: $v_{max} = 5 \,m \,s^{-1}$ and $a_{max} = 10 \,m \,s^{-2}$.Dividing the expression for $a_{max}$ by $v_{max}$, we get: $\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega$.Therefore, $\omega = \frac{10}{5} = 2 \,rad \,s^{-1}$.The time period $T$ is related to the angular frequency by the formula $T = \frac{2\pi}{\omega}$.Substituting the value of $\omega$, we get $T = \frac{2\pi}{2} = \pi \,s$.

If the maximum velocity and maximum acceleration of a particle executing simple harmonic motion are respectively $5 \,m \,s^{-1}$ and $10 \,m \,s^{-2}$, then the time period of oscillation of the particle is

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