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A velocity-displacement graph for a particle performing SHM is a standard ellipse with maximum velocity v_max = 10 cm/s at the mean position (x = 0) and maximum displacement A = 2.5 cm. Which of the following statements are correct?

1. The time period of the particle is 1.57 s.
2. The maximum acceleration will be 40 cm/s².
3. The velocity of the particle is 2*sqrt(21) cm/s when it is at a distance 1 cm from the mean position.
4. The minimum acceleration is 4 cm/s².

Correct answer: The time period of the particle is 1.57 s.

Solution

From the v-x graph, the intercepts give v_max = 10 cm/s at x = 0 and A = 2.5 cm. So w = v_max / A = 10 / 2.5 = 4 rad/s. Time period T = 2*pi / w = 2*pi / 4 = pi/2 ≈ 1.57 s (option A: correct). Maximum acceleration = w²*A = 16 * 2.5 = 40 cm/s² (option B: correct). At x = 1 cm: v = w*sqrt(A² - x²) = 4*sqrt(6.25 - 1) = 4*sqrt(5.25) = 4*sqrt(21/4) = 2*sqrt(21) cm/s (option C: correct). Minimum acceleration in SHM is zero (at mean position x = 0); option D claims 4 cm/s² which is incorrect.

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