MHT CET 2026 17th April Evening Shift | Simple Harmonic Motion Question 17 | Physics

Two simple pendulums of lengths $L_1$ and $L_2$ have periodic time $T_1$ and $T_2$ respectively $(T_1 > T_2)$. The time period of the pendulum of length $(L_1-L_2)$ is
$[(L_1-L_2) > 60\text{ cm}]$

$\sqrt{T_1^2 + T_2^2}$

$\sqrt{T_1^2 - T_2^2}$

$T_1 + T_2$

$T_1 - T_2$

MHT CET 2026 17th April Morning Shift

MCQ (Single Correct Answer)

-0

For a particle executing S.H.M., its potential energy is 8 times its kinetic energy at a certain displacement '$x$' from the mean position. If '$A$' is the amplitude of S.H.M., the value of '$x$' is

$\dfrac{2}{\sqrt{3}}A$

$\dfrac{\sqrt{2}}{3}A$

$\dfrac{2\sqrt{2}}{3}A$

$\dfrac{3}{\sqrt{2}}A$

MHT CET 2026 17th April Morning Shift

MCQ (Single Correct Answer)

-0

A particle executing simple harmonic motion starts from mean position with amplitude '$A$' and periodic time '$T$'. At what displacement is its speed one-fourth of the maximum speed?

$\dfrac{A}{\sqrt{15}}$

$\dfrac{A}{4}$

$\dfrac{4A}{\sqrt{15}}$

$\dfrac{A\sqrt{15}}{4}$

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