Question

For a body in S.H.M. the velocity is given by the relation $V=\sqrt{144-16 \mathrm{x}^{2}} \mathbf{m s}^{-1}$. The maximum acceleration is

• A. 12 \mathrm{~m} / \mathrm{s}^{2}
• B. 16 \mathrm{~m} / \mathrm{s}^{2}
• C. 36 \mathrm{~m} / \mathrm{s}^{2}
• D. 48 \mathrm{~m} / \mathrm{s}^{2}

Answer 1

Answer: (D)

48 \mathrm{~m} / \mathrm{s}^{2}

Answer 2

The maximum acceleration in S.H.M. can be found using the formula $a_{max} = \omega^2 A$, where $A$ is the amplitude. From the given velocity equation, we can deduce that the maximum displacement (amplitude) is 3 m, and the angular frequency $\omega$ can be derived from the velocity equation. The maximum acceleration is then calculated as $16 \times 3 = 48 \mathrm{~m/s^2}$.