1
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
A body of mass $0.4$ kg performs simple harmonic motion. It experiences a restoring force of $0.4$ N when its displacement from the mean position is 4 cm. The force constant and magnitude of acceleration respectively are
A
$5\ \text{N/}_\text{m}, 0.5\ \text{m/}_{\text{s}^2}$
B
$10\ \text{N/}_\text{m}, 1\ \text{m/}_{\text{s}^2}$
C
$15\ \text{N/}_\text{m}, 1.5\ \text{m/}_{\text{s}^2}$
D
$20\ \text{N/}_\text{m}, 2\ \text{m/}_{\text{s}^2}$
2
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
The minimum phase difference between two simple harmonic motions is
$x_1 = \dfrac{1}{\sqrt{2}} \sin \omega t + \dfrac{1}{\sqrt{2}} \cos \omega t$
$x_2 = \sin \omega t + \cos \omega t$     $\left[ \sin \dfrac{\pi}{4} = \cos \dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}} \right]$
A
zero
B
$\dfrac{\pi^C}{3}$
C
$\dfrac{\pi^C}{4}$
D
$\dfrac{\pi^C}{5}$
3
MHT CET 2026 15th April Evening Shift
MCQ (Single Correct Answer)
+1
-0
A pendulum clock is running slow, In order to correct it, we should
A
reduce the amplitude of oscillation.
B
reduce the mass of the bob.
C
reduce the length of pendulum.
D
increase the length of pendulum.
4
MHT CET 2026 15th April Morning Shift
MCQ (Single Correct Answer)
+1
-0
A particle executes linear S.H.M. with amplitude $4$ cm. The magnitude of velocity and acceleration is equal when it is at $3$ cm from mean position. Time period is
A
$\dfrac{3\pi}{\sqrt{2}}$ s
B
$\dfrac{6\pi}{\sqrt{7}}$ s
C
$\dfrac{2\pi}{\sqrt{7}}$ s
D
$\dfrac{4\pi}{\sqrt{7}}$ s

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