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Question

The displacement of the particle varies with time x = 12 sin $$\omega$$ t - 16 $$sin^{3}$$ $$\omega$$ t. If its motion is SHM, then maximum acceleration is


A
12 ω2
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B
36 ω2
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C
144 ω2
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D
192 ω2
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Solution

The correct option is B 36 $$\omega^{2}$$
$$x=12\sin(\omega t)-16\sin^3(\omega t)$$

$$x=4(3\sin(\omega t)-4\sin^3(\omega t)$$

$$x=4\sin(3\omega t)$$

Maximum acceleration$$=A\omega^2$$

$$A=4,\ \omega=3\omega$$

Max acceleration$$=4(3\omega)^2=36\omega^2$$

Option $$\textbf B$$ is the correct answer.

Physics

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