MHT CET 2025 25th April Morning Shift | Simple Harmonic Motion Question 9 | Physics

The displacement of particle in S.H.M. is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 6)$. Its speed will be maximum at time $\left(\sin 90^{\circ}=1\right)$

$\frac{\pi}{3 \omega} \mathrm{~s}$

$\frac{\pi}{2 \omega} \mathrm{~s}$

$\frac{\pi}{\omega} \mathrm{s}$

$\frac{\pi}{4 \omega} \mathrm{~s}$

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

An object of mass 0.2 kg executes simple harmonic oscillations along the x -axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy 1 J and potential energy 0.6 J . The amplitude of oscillation is

0.06 m

0.6 m

0.08 m

0.8 m

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

The motion of the particle is given by the equation $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$.

The motion of the particle is

simple harmonic with amplitude $(\mathrm{A}+\mathrm{B})$

simple harmonic with amplitude $(A-B)$

simple harmonic with amplitude $\left(A^2+B^2\right)^{\frac{1}{2}}$

not simple harmonic

MHT CET 2025 23rd April Evening Shift

MCQ (Single Correct Answer)

-0

A particle is executing S.H.M. of amplitude ' $A$ '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is

$\pm \frac{\mathrm{A}}{4}$

$\pm \frac{A}{2}$

$\pm \frac{\mathrm{A}}{\sqrt{3}}$

$$ \pm \frac{A}{\sqrt{2}} $$

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