MHT CET 2025 21st April Morning Shift | Simple Harmonic Motion Question 13 | Physics

A simple pendulum has time period ' $\mathrm{T}_1$ '. The point of suspension is now moved upward according to equation $\mathrm{y}=\mathrm{kt}^2$ where $\mathrm{k}=1 \mathrm{~m} / \mathrm{s}^2$. If new time period is ' $\mathrm{T}_2$ ' then $\mathrm{T}_1^2 / \mathrm{T}_2^2$ will be ( $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

$2 / 3$

$5 / 6$

$6 / 5$

$3 / 2$

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

A mass attached to a spring performs S.H.M. whose displacement is $\mathrm{x}=3 \times 10^{-3} \cos 2 \pi \mathrm{t}$ metre. The time taken to obtain maximum speed for the first time is

$\frac{1}{12} \mathrm{~s}$

$\frac{1}{8} \mathrm{~s}$

$\frac{1}{4} \mathrm{~s}$

$\frac{1}{2} \mathrm{~s}$

MHT CET 2025 20th April Evening Shift

MCQ (Single Correct Answer)

-0

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)

-0

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

$7: 9$

$5: 4$

$3: 2$

$8: 7$

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