MHT CET 2026 16th April Evening Shift | Simple Harmonic Motion Question 21 | Physics

A mass $m_1$ performs S.H.M. with amplitude A which is connected to horizontal spring. While mass $m_1$ passing through mean position, another mass $m_2$ ($m_2 < m_1$) is placed on it so that both masses move together with amplitude $A_1$. The ratio of $A/A_1$ is

$\left[\dfrac{m_1 + m_2}{m_1}\right]^{\frac{1}{2}}$

$\left[\dfrac{m_1}{m_1 + m_2}\right]^{\frac{1}{2}}$

$\dfrac{m_1}{m_1 + m_2}$

$\dfrac{m_2}{m_1 + m_2}$

MHT CET 2026 16th April Evening Shift

MCQ (Single Correct Answer)

-0

The maximum potential energy of a block executing S.H.M. is E and amplitude of oscillation is 'A'. The kinetic energy of the block at amplitude A/2 is

$\dfrac{3\text{E}}{4}$

$\text{E}$

$\dfrac{2\text{E}}{4}$

$\dfrac{5\text{E}}{4}$

MHT CET 2026 16th April Morning Shift

MCQ (Single Correct Answer)

-0

A rectangular block of mass 'm' and cross-sectional area 'A' floats on a liquid of density '$\varrho$'. It is given a small vertical displacement from equilibrium, it starts oscillating with frequency (g=acceleration due to gravity)

$2\pi\sqrt{\dfrac{\text{m}}{\text{A}\varrho\text{g}}}$

$2\pi\sqrt{\dfrac{\text{A}\varrho\text{g}}{\text{m}}}$

$\dfrac{1}{2\pi}\sqrt{\dfrac{\text{A}\varrho\text{g}}{\text{m}}}$

$\dfrac{1}{2\pi}\sqrt{\dfrac{\text{m}}{\text{A}\varrho\text{g}}}$

MHT CET 2026 16th April Morning Shift

MCQ (Single Correct Answer)

-0

For a particle P performing S.H.M., when displacement is 'x', potential energy and restoring force acting on it is denoted by 'E' and 'F' respectively. The relation between x, E and F is

$\dfrac{2\text{E}}{\text{F}} - x = 0$

$\dfrac{2\text{E}}{\text{F}} + x = 0$

$\dfrac{2\text{F}}{\text{x}} + \text{F} = 0$

$\dfrac{2\text{E}}{\text{x}} - \text{F} = 0$

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