MHT CET 2026 16th April Evening Shift | Simple Harmonic Motion Question 22 | Physics

The maximum potential energy of a block executing S.H.M. is E and amplitude of oscillation is 'A'. The kinetic energy of the block at amplitude A/2 is

$\dfrac{3\text{E}}{4}$

$\text{E}$

$\dfrac{2\text{E}}{4}$

$\dfrac{5\text{E}}{4}$

MHT CET 2026 16th April Morning Shift

MCQ (Single Correct Answer)

-0

A rectangular block of mass 'm' and cross-sectional area 'A' floats on a liquid of density '$\varrho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency (g=acceleration due to gravity)

$2\pi\sqrt{\dfrac{\text{m}}{\text{A}\varrho\text{g}}}$

$2\pi\sqrt{\dfrac{\text{A}\varrho\text{g}}{\text{m}}}$

$\dfrac{1}{2\pi}\sqrt{\dfrac{\text{A}\varrho\text{g}}{\text{m}}}$

$\dfrac{1}{2\pi}\sqrt{\dfrac{\text{m}}{\text{A}\varrho\text{g}}}$

MHT CET 2026 16th April Morning Shift

MCQ (Single Correct Answer)

-0

For a particle P performing S.H.M., when displacement is 'x', potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between x, E and F is

$\dfrac{2\text{E}}{\text{F}} - x = 0$

$\dfrac{2\text{E}}{\text{F}} + x = 0$

$\dfrac{2\text{F}}{\text{x}} + \text{F} = 0$

$\dfrac{2\text{E}}{\text{x}} - \text{F} = 0$

MHT CET 2026 15th April Evening Shift

MCQ (Single Correct Answer)

-0

A body of mass $0.4$ kg performs simple harmonic motion. It experiences a restoring force of $0.4$ N when its displacement from the mean position is 4 cm. The force constant and magnitude of acceleration respectively are

$5\ \text{N/}_\text{m}, 0.5\ \text{m/}_{\text{s}^2}$

$10\ \text{N/}_\text{m}, 1\ \text{m/}_{\text{s}^2}$

$15\ \text{N/}_\text{m}, 1.5\ \text{m/}_{\text{s}^2}$

$20\ \text{N/}_\text{m}, 2\ \text{m/}_{\text{s}^2}$

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