MHT CET 2023 14th May Evening Shift | Simple Harmonic Motion Question 3 | Physics

A block of mass '$$M$$' rests on a piston executing S.H.M. of period one second. The amplitude of oscillations, so that the mass is separated from the piston, is (acceleration due to gravity, $$\mathrm{g}=10 \mathrm{~ms}^{-2}, \pi^2=10$$ )

0.25 m

0.5 m

1 m

$$\infty$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

A simple pendulum of length '$$l$$' and a bob of mass '$$\mathrm{m}$$' is executing S.H.M. of small amplitude '$$A$$'. The maximum tension in the string will be ($$\mathrm{g}=$$ acceleration due to gravity)

$$2 \mathrm{~mg}$$

$$\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)^2\right]$$

$$\mathrm{mg}\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]^2$$

$$m g\left[1+\left(\frac{\mathrm{A}}{\ell}\right)\right]$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

The displacement of a particle executing S.H.M. is $$x=\mathrm{a} \sin (\omega t-\phi)$$. Velocity of the particle at time $$\mathrm{t}=\frac{\phi}{\omega}$$ is $$\left(\cos 0^{\circ}=1\right)$$

$$\omega \cos \phi$$

$$\mathrm{a} \omega$$

$$\omega \cos 2 \phi$$

$$-\mathrm{a} \omega \cos 2 \phi$$

MHT CET 2023 14th May Morning Shift

MCQ (Single Correct Answer)

-0

The bob of simple pendulum of length '$$L$$' is released from a position of small angular displacement $$\theta$$. Its linear displacement at time '$$\mathrm{t}$$' is ( $$\mathrm{g}=$$ acceleration due to gravity)

$$L \theta \cos \left[\sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \sin \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \cos \left[2 \pi \sqrt{\frac{g}{L}} \cdot t\right]$$

$$L \theta \sin \left[\sqrt{\frac{g}{L}} \cdot t\right]$$

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