MHT CET 2020 16th October Evening Shift | Simple Harmonic Motion Question 121 | Physics

A particle performs simple harmonic motion with period of 3 s . The time taken by it to cover a distance equal to half the amplitude from mean position is [$$\sin 30^{\circ}=0.5$$]

$$\frac{1}{4} \mathrm{~s}$$

$$\frac{3}{2} \mathrm{~s}$$

$$\frac{3}{4} \mathrm{~s}$$

$$\frac{1}{2} \mathrm{~s}$$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

A simple pendulum of length $$I$$ has a bob of mass $$m$$. It executes SHM of small amplitude A. The maximum tension in the string is ($$g=$$ acceleration due to gravity)

$$m g$$

$$m g\left(\frac{A^2}{I^2}+1\right)$$

2 mg

$$m g\left(\frac{A}{I}+1\right)$$

MHT CET 2020 16th October Evening Shift

MCQ (Single Correct Answer)

-0

A block of mass $$m$$ attached to one end of the vertical spring produces extension $$x$$. If the block is pulled and released, the periodic time of oscillation is

$$2 \pi \sqrt{\frac{x}{4 g}}$$

$$2 \pi \sqrt{\frac{2 x}{g}}$$

$$2 \pi \sqrt{\frac{x}{2 g}}$$

$$2 \pi \sqrt{\frac{x}{g}}$$

MHT CET 2020 16th October Morning Shift

MCQ (Single Correct Answer)

-0

A simple pendulum of length $$L$$ has mass $$m$$ and it oscillates freely with amplitude $$A$$. At extreme position, its potential energy is ($$g=$$ acceleration due to gravity)

$$\frac{m g A}{L}$$

$$\frac{m g A}{2 l}$$

$$\frac{m g A^2}{L}$$

$$\frac{m g A^2}{2 L}$$

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