JEE Main 2024 (Online) 4th April Evening Shift | Simple Harmonic Motion Question 6 | Physics

In simple harmonic motion, the total mechanical energy of given system is $$E$$. If mass of oscillating particle $$P$$ is doubled then the new energy of the system for same amplitude is:

JEE Main 2024 (Online) 4th April Evening Shift Physics - Simple Harmonic Motion Question 4 English

$$E / \sqrt{2}$$

$$2 E$$

$$E \sqrt{2}$$

$$E$$

JEE Main 2024 (Online) 1st February Morning Shift

MCQ (Single Correct Answer)

-1

A simple pendulum of length $1 \mathrm{~m}$ has a wooden bob of mass $1 \mathrm{~kg}$. It is struck by a bullet of mass $10^{-2} \mathrm{~kg}$ moving with a speed of $2 \times 10^2 \mathrm{~ms}^{-1}$. The bullet gets embedded into the bob. The height to which the bob rises before swinging back is. (use $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

$0.20 \mathrm{~m}$

$0.40 \mathrm{~m}$

$0.30 \mathrm{~m}$

$0.35 \mathrm{~m}$

JEE Main 2024 (Online) 29th January Evening Shift

MCQ (Single Correct Answer)

-1

The bob of a pendulum was released from a horizontal position. The length of the pendulum is $$10 \mathrm{~m}$$. If it dissipates $$10 \%$$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

[Use, $$\mathrm{g}: 10 \mathrm{~ms}^{-2}$$]

$$5 \sqrt{6} \mathrm{~ms}^{-1}$$

$$5 \sqrt{5} \mathrm{~ms}^{-1}$$

$$2 \sqrt{5} \mathrm{~ms}^{-1}$$

$$6 \sqrt{5} \mathrm{~ms}^{-1}$$

JEE Main 2024 (Online) 27th January Evening Shift

MCQ (Single Correct Answer)

-1

A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $$(\theta)$$ of thread deflection in the extreme position will be :

$$\tan ^{-1}\left(\frac{1}{2}\right)$$

$$2 \tan ^{-1}\left(\frac{1}{2}\right)$$

$$2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)$$

$$\tan ^{-1}(\sqrt{2})$$

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