MHT CET 2026 15th April Morning Shift | Simple Harmonic Motion Question 29 | Physics

A particle starts from mean position and performs S.H.M. with period $6$ second. At what time its kinetic energy is $50\%$ of total energy? ($\cos 45^\circ = 1/\sqrt{2}$)

$0.75$ s

$0.50$ s

$0.25$ s

$3$ s

MHT CET 2026 15th April Morning Shift

MCQ (Single Correct Answer)

-0

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is $1.6$ m. The period of Oscillation of the sphere is (Acceleration due to gravity $g = 10 \text{ m/s}^2$)

$0.2\pi$

$0.4\pi$

$0.8\pi$

$\pi$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

All the springs in fig. (a), (b) and (c) are identical, each having force constant K each. Mass m is attached to each system. If $\mathrm{T}_a, \mathrm{~T}_b$ and $\mathrm{T}_{\mathrm{c}}$ are the time periods of oscillations of the three systems in fig. (a), (b) and (c) respectively, then

MHT CET 2025 5th May Evening Shift Physics - Simple Harmonic Motion Question 31 English

$\quad \mathrm{T}_{\mathrm{a}}=\sqrt{2} \mathrm{~T}_{\mathrm{b}}$

$\quad \mathrm{T}_{\mathrm{a}}=\frac{\mathrm{T}_{\mathrm{c}}}{\sqrt{2}}$

$\mathrm{T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{a}}$

$\quad \mathrm{T}_{\mathrm{b}}=2 \mathrm{~T}_{\mathrm{c}}$

MHT CET 2025 5th May Evening Shift

MCQ (Single Correct Answer)

-0

A point particle of mass 200 gram is executing S.H.M. of amplitude 0.2 m . When the particle passes through the mean position, its kinetic energy is $16 \times 10^{-3} \mathrm{~J}$. The equation of motion of this particle is (Initial phase of oscillation $=0^{\circ}$ )

$\mathrm{Y}=0.2 \sin (4 \mathrm{t})$

$\mathrm{Y}=0.2 \sin \left(\frac{\mathrm{t}}{4}\right)$

$\mathrm{Y}=0.2 \sin \left(\frac{\mathrm{t}}{2}\right)$

$\mathrm{Y}=0.2 \sin (2 \mathrm{t})$

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