1
JEE Main 2024 (Online) 29th January Evening Shift
Numerical
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A simple harmonic oscillator has an amplitude $$A$$ and time period $$6 \pi$$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $$x=$$ A to $$x=\frac{\sqrt{3}}{2}$$ A will be $$\frac{\pi}{x} \mathrm{~s}$$, where $$x=$$ _________.

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2
JEE Main 2024 (Online) 29th January Morning Shift
Numerical
+4
-1
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When the displacement of a simple harmonic oscillator is one third of its amplitude, the ratio of total energy to the kinetic energy is $$\frac{x}{8}$$, where $$x=$$ _________.

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3
JEE Main 2024 (Online) 27th January Morning Shift
Numerical
+4
-1
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A particle executes simple harmonic motion with an amplitude of $$4 \mathrm{~cm}$$. At the mean position, velocity of the particle is $$10 \mathrm{~cm} / \mathrm{s}$$. The distance of the particle from the mean position when its speed becomes $$5 \mathrm{~cm} / \mathrm{s}$$ is $$\sqrt{\alpha} \mathrm{~cm}$$, where $$\alpha=$$ ________.

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4
JEE Main 2023 (Online) 13th April Morning Shift
Numerical
+4
-1
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At a given point of time the value of displacement of a simple harmonic oscillator is given as $$\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)$$. If amplitude is $$40 \mathrm{~cm}$$ and kinetic energy at that time is $$200 \mathrm{~J}$$, the value of force constant is $$1.0 \times 10^{x} ~\mathrm{Nm}^{-1}$$. The value of $$x$$ is ____________.

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