JEE Main 2017 · 8 Apr · Shift 1 · Physics Q12

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The ratio of maximum acceleration to maximum velocity in a simple harmonic motion is 10 s $^{-1}$ . At, t = 0 the displacement is 5 m. What is the maximum acceleration ? The initial phase is $\dfrac\pi4$ .

500 m/s

^2

500\sqrt2

m/s

^2

750 m/s

^2

750\sqrt2

m/s

^2

Answer and worked solution

Official answer: B

$\dfrac{a_{max}}{v_{max}}=\omega=10$ s⁻¹. From $x=A\sin\left(\omega t+\frac\pi4\right)$ , $5=\dfrac A{\sqrt2}$ , so $A=5\sqrt2$ m and
$a_{max}=\omega^2A=500\sqrt2\text{ m/s}^2.$

Topics:Simple harmonic motion

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