MHT CET 2024 16th May Evening Shift | Simple Harmonic Motion Question 1 | Physics

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

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

$2 \pi \mathrm{~s}$

$20 \pi \mathrm{~s}$

$5 \pi \mathrm{~s}$

MHT CET 2024 16th May Evening Shift

MCQ (Single Correct Answer)

-0

A particle performs linear S.H.M. At a particular instant, velocity of the particle is ' $u$ ' and acceleration is ' $\alpha$ ' while at another instant, velocity is ' $v$ ' and acceleration is ' $\beta$ ' $(0<\alpha<\beta)$. The distance between the two positions is

$\frac{u^2-v^2}{\alpha+\beta}$

$\frac{u^2+v^2}{\alpha+\beta}$

$\frac{u^2-v^2}{\alpha-\beta}$

$\frac{u^2+v^2}{\alpha-\beta}$

MHT CET 2024 16th May Morning Shift

MCQ (Single Correct Answer)

-0

A particle executing S.H.M. has velocities ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' at distances ' $x_1$ ' and ' $x_2$ ' respectively, from the mean position. Its frequency is

$\frac{1}{2 \pi} \sqrt{\frac{V_1^2-V_2^2}{x_1^2-x_2^2}}$

$2 \pi \sqrt{\frac{x_1^2-x_2^2}{v_1^2-V_2^2}}$

$\frac{1}{2 \pi} \sqrt{\frac{V_2^2-V_1^2}{x_1^2-x_2^2}}$

$2 \pi \sqrt{\frac{\mathrm{x}_1^2-\mathrm{x}_2^2}{\mathrm{~V}_2^2-\mathrm{V}_1^2}}$

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