MHT CET 2025 19th April Morning Shift | Simple Harmonic Motion Question 5 | Physics

A mass suspended from a vertical spring performs S.H.M. of period 0.1 second. The spring is unstretched at the highest point of suspension. Maximum speed of the mass is (Gravitational acceleration $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

$\frac{1}{2 \pi} \mathrm{~m} / \mathrm{s}$

$\frac{1}{\pi} \mathrm{~m} / \mathrm{s}$

$\frac{2}{\pi} \mathrm{~m} / \mathrm{s}$

$\pi \mathrm{m} / \mathrm{s}$

MHT CET 2025 19th April Morning Shift

MCQ (Single Correct Answer)

-0

' $P$ ' and ' $Q$ ' are fixed points in same plane and mass ' $m$ ' is tied by string as shown in figure. If the mass is displaced slightly out of this plane and released, it will oscillate with time period $(\mathrm{PQ}=2 \mathrm{~d}, \mathrm{PR}=\mathrm{QR}=\mathrm{L})(\mathrm{g}=$ gravitational acceleration)

MHT CET 2025 19th April Morning Shift Physics - Simple Harmonic Motion Question 2 English

$2 \pi \sqrt{\frac{\mathrm{~L}}{\mathrm{~g}}}$

$2 \pi \sqrt{\frac{\mathrm{~L}^2}{\mathrm{~g}}}$

$2 \pi \sqrt{\frac{\left(L^2-d^2\right)^{1 / 2}}{g}}$

$2 \pi \sqrt{\frac{\left(L^2+d^2\right)^{1 / 2}}{g}}$

MHT CET 2024 16th May Evening Shift

MCQ (Single Correct Answer)

-0

The bob of a pendulum of length ' $l$ ' is pulled aside from its equilibrium position through an angle ' $\theta$ ' and then released. The bob will then pass through its equilibrium position with speed ' $v$ ', where ' $v$ ' equal to ( $g=$ acceleration due to gravity)

$\sqrt{2 g l(1-\cos \theta)}$

$\sqrt{2 \mathrm{~g} l(1+\sin \theta)}$

$\sqrt{2 g l(1-\sin \theta)}$

$\sqrt{2 \mathrm{~g} l(1+\cos \theta)}$

MHT CET 2024 16th May Evening Shift

MCQ (Single Correct Answer)

-0

The kinetic energy of a particle, executing simple harmonic motion is 16 J when it is in mean position. If amplitude of motion is 25 cm and the mass of the particle is 5.12 kg , the period of oscillation is

$\frac{\pi}{5} \mathrm{~s}$

$2 \pi \mathrm{~s}$

$20 \pi \mathrm{~s}$

$5 \pi \mathrm{~s}$

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