1
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+1
-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

A
0.06 m
B
0.6 m
C
0.08 m
D
0.8 m
2
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+1
-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

A
simple harmonic with amplitude $(\mathrm{A}+\mathrm{B})$
B
simple harmonic with amplitude $(A-B)$
C
simple harmonic with amplitude $\left(A^2+B^2\right)^{\frac{1}{2}}$
D
not simple harmonic
3
MHT CET 2025 23rd April Evening Shift
MCQ (Single Correct Answer)
+1
-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

A
$\pm \frac{\mathrm{A}}{4}$
B
$\pm \frac{A}{2}$
C
$\pm \frac{\mathrm{A}}{\sqrt{3}}$
D

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

4
MHT CET 2025 23rd April Morning Shift
MCQ (Single Correct Answer)
+1
-0

A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is

A
$2: 1$
B
$3: 1$
C
$4: 1$
D
$8: 1$
MHT CET Subjects
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