MHT CET 2024 9th May Evening Shift | Simple Harmonic Motion Question 21 | Physics

A particle is executing a linear simple harmonic motion. Let ' $\mathrm{V}_1$ ' and ' $\mathrm{V}_2$ ' are its speed at distance ' $x_1$ ' and ' $x_2$ ' from the equilibrium position. The amplitude of oscillation is

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

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

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

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

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

In S.H.M. the displacement of a particle at an instant is $Y=A \cos 30^{\circ}$, where $A=40 \mathrm{~cm}$ and kinetic energy is 200 J . If force constant is $1 \times 10^{\times} \mathrm{N} / \mathrm{m}$, then x will be $\left(\cos 30^{\circ}=\sqrt{3} / 2\right)$

MHT CET 2024 9th May Morning Shift

MCQ (Single Correct Answer)

-0

A particle is performing S.H.M. with an amplitude 4 cm . At the mean position the velocity of the particle is $12 \mathrm{~cm} / \mathrm{s}$. When the speed of the particle becomes $6 \mathrm{~cm} / \mathrm{s}$, the distance of the particle from mean position is

$\sqrt{3} \mathrm{~cm}$

$\sqrt{6} \mathrm{~cm}$

$2 \sqrt{3} \mathrm{~cm}$

$2 \sqrt{6} \mathrm{~cm}$

MHT CET 2024 4th May Evening Shift

MCQ (Single Correct Answer)

-0

The maximum velocity and maximum acceleration of a particle performing a linear S.H.M. is ' $\alpha$ ' and ' $\beta$ ' respectively. Then the path length of the particle is

$\frac{\alpha^2}{\beta}$

$\frac{\beta \alpha^2}{2 \alpha^2}$

$\frac{2 \alpha^2}{\beta}$

$\frac{2 \beta}{\alpha^2}$

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