MHT CET 2025 20th April Evening Shift | Simple Harmonic Motion Question 15 | Physics

A particle executes a linear S.H.M. In two of its positions the velocities are $\mathrm{V}_1, \mathrm{~V}_2$ and accelerations are $\mathrm{a}_1$ and $\mathrm{a}_2$ respectively $\left(0

$\frac{V_1^2-V_2^2}{a_1-a_2}$

$\frac{V_2^2-V_1^2}{a_1-a_2}$

$\frac{V_1^2-V_2^2}{a_1+a_2}$

$\frac{v_2^2-v_1^2}{\left(a_1^2+a_2^2\right)}$

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

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A mass $x$ gram is suspended from a light spring. It is pulled in downward direction and released so that mass performs S.H.M. of period T. If mass is increased by Y gram, the period becomes $\frac{4 \mathrm{~T}}{3}$. The ratio of $\mathrm{Y} / \mathrm{x}$ is

$7: 9$

$5: 4$

$3: 2$

$8: 7$

MHT CET 2025 20th April Morning Shift

MCQ (Single Correct Answer)

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The period of S. H.M. of a particle is 16 second. The phase difference between the positions at $\mathrm{t}=2 \mathrm{~s}$ and $\mathrm{t}=4 \mathrm{~s}$ will be

$\pi$

$\frac{\pi}{2}$

$\frac{\pi}{4}$

$\frac{\pi}{8}$

MHT CET 2025 20th April Morning Shift

MCQ (Single Correct Answer)

-0

If the period of a oscillation of mass ' m ' suspended from a spring is 2 s , then the period of suspended mass ' 4 m ' with the same spring will be

1 s

3 s

2 s

4 s

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